Velocity of Simple Harmonic Motion Questions in English

(A) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$.
At displacement $x = \frac{2A}{3}$,the velocity is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}A\omega}{3}$.
When the speed is increased to three times,the new velocity $v' = 3v = 3 \times \frac{\sqrt{5}A\omega}{3} = \sqrt{5}A\omega$.
Let the new amplitude be $A'$. The new velocity at the same position $x = \frac{2A}{3}$ is $v' = \omega \sqrt{(A')^2 - x^2}$.
Equating the expressions: $\sqrt{5}A\omega = \omega \sqrt{(A')^2 - (\frac{2A}{3})^2}$.
Squaring both sides: $5A^2 = (A')^2 - \frac{4A^2}{9}$.
$(A')^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$.
Taking the square root: $A' = \frac{7A}{3}$.
Comparing this with $\frac{nA}{3}$,we get $n = 7$.

(A) Given,the displacement equation is $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$.
The time period $T = 3.14 \ s$. We know that $T = \frac{2\pi}{\omega}$.
Taking $\pi \approx 3.14$,we have $3.14 = \frac{2 \times 3.14}{\omega}$,which gives $\omega = 2 \ rad/s$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt} \left[10 \sin \left(\omega t + \frac{\pi}{3}\right)\right]$.
$v = 10 \omega \cos \left(\omega t + \frac{\pi}{3}\right)$.
At $t = 0$,the velocity is $v = 10 \times 2 \times \cos \left(0 + \frac{\pi}{3}\right)$.
$v = 20 \times \cos \left(\frac{\pi}{3}\right) = 20 \times \frac{1}{2} = 10 \ m/s$.

(A) The maximum velocity $(V_{\max})$ of a particle in simple harmonic motion is given by the formula: $V_{\max} = \omega A$.
Here, $\omega$ is the angular frequency and $A$ is the amplitude.
We know that $\omega = \frac{2\pi}{T}$, where $T$ is the time period.
Given: $A = 0.06 \,m$ and $T = 3.14 \,s \approx \pi \,s$.
Substituting the values:
$V_{\max} = \left(\frac{2\pi}{\pi}\right) \times 0.06 = 2 \times 0.06 = 0.12 \,m/s$.
To convert the velocity into $cm/s$, we multiply by $100$:
$V_{\max} = 0.12 \times 100 = 12 \,cm/s$.

(A) Given: Position $x = 4 \ m$,Velocity $v = 2 \ ms^{-1}$,Acceleration $a = 16 \ ms^{-2}$.
The magnitude of acceleration in simple harmonic motion is given by $|a| = \omega^2 x$.
Substituting the values: $16 = \omega^2(4) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2 \ rad/s$.
The velocity in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude.
Squaring both sides: $v^2 = \omega^2(A^2 - x^2) \Rightarrow \frac{v^2}{\omega^2} = A^2 - x^2$.
Rearranging for $A$: $A^2 = \frac{v^2}{\omega^2} + x^2$.
Substituting the values: $A^2 = \frac{2^2}{2^2} + 4^2 = \frac{4}{4} + 16 = 1 + 16 = 17$.
Therefore,$A = \sqrt{17} \ m$. Comparing this with $\sqrt{x} \ m$,we get $x = 17$.

(C) For a particle in $S.H.M.$,the velocity $V$ at displacement $x$ is given by $V^2 = \omega^2(A^2 - x^2)$,and acceleration $a$ is given by $a = -\omega^2 x$.
From the acceleration equation,we have $x = -a/\omega^2$.
Substituting this into the velocity equation: $V^2 = \omega^2 A^2 - \omega^2 x^2 = \omega^2 A^2 - \omega^2 (-a/\omega^2)^2 = \omega^2 A^2 - a^2/\omega^2$.
For two positions with velocities $V_1, V_2$ and accelerations $a_1, a_2$:
$V_1^2 = \omega^2 A^2 - a_1^2/\omega^2$
$V_2^2 = \omega^2 A^2 - a_2^2/\omega^2$
Subtracting the two equations: $V_1^2 - V_2^2 = (a_2^2 - a_1^2)/\omega^2$.
Since $a = -\omega^2 x$,the distance between two positions $x_1$ and $x_2$ is $d = |x_1 - x_2| = |(-a_1/\omega^2) - (-a_2/\omega^2)| = |a_2 - a_1|/\omega^2$.
From the subtraction result,$1/\omega^2 = (V_1^2 - V_2^2) / (a_2^2 - a_1^2)$.
Substituting this into the distance formula: $d = |a_2 - a_1| \cdot \frac{V_1^2 - V_2^2}{a_2^2 - a_1^2} = \frac{V_1^2 - V_2^2}{a_2 + a_1}$ (since $a_2 > a_1$).
Thus,the correct option is $C$.

(C) The displacement of the particle is given by $x = A \cos(\omega t)$,where $A = 3 \times 10^{-3} \text{ m}$ and $\omega = 2 \pi \text{ rad/s}$.
The velocity $v$ of the particle is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -A \omega \sin(\omega t)$.
The speed is $|v| = A \omega |\sin(\omega t)|$.
Maximum speed occurs when $|\sin(\omega t)| = 1$,which means $\omega t = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$.
For the first time,$\omega t = \frac{\pi}{2}$.
Substituting $\omega = 2 \pi$,we get $2 \pi t = \frac{\pi}{2}$.
Solving for $t$,we get $t = \frac{1}{4} \text{ s}$.

(D) The speed of a particle in $S.H.M.$ at a displacement '$x$' is given by $u = \omega \sqrt{a^2 - x^2}$.
The maximum speed of the particle is $V_{\text{max}} = a\omega$.
Given that the speed '$u$' is half of the maximum speed,we have $u = \frac{V_{\text{max}}}{2} = \frac{a\omega}{2}$.
Equating the two expressions for speed: $\frac{a\omega}{2} = \omega \sqrt{a^2 - x^2}$.
Canceling $\omega$ from both sides: $\frac{a}{2} = \sqrt{a^2 - x^2}$.
Squaring both sides: $\frac{a^2}{4} = a^2 - x^2$.
Rearranging for $x^2$: $x^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}$.
Taking the square root: $x = \frac{\sqrt{3}a}{2}$.

(A) The displacement of the particle is $x = A \cos(\omega t + \pi/6)$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -A \omega \sin(\omega t + \pi/6)$.
The speed is maximum when the magnitude of velocity $|v|$ is maximum,which occurs when $|\sin(\omega t + \pi/6)| = 1$.
This happens when the argument $(\omega t + \pi/6) = \pi/2$ (or $3\pi/2$,etc.).
Setting $\omega t + \pi/6 = \pi/2$,we get $\omega t = \pi/2 - \pi/6 = 3\pi/6 - \pi/6 = 2\pi/6 = \pi/3$.
Therefore,$t = \frac{\pi}{3 \omega} \text{ s}$.

(B) Given: Mass $m = 1 \,kg$, Force constant $k = 4 \,N/m$, Velocity $v = 20 \,cm/s = 0.2 \,m/s$, Amplitude $A = 0.4 \,m$.
First, calculate the angular frequency $\omega = \sqrt{k/m} = \sqrt{4/1} = 2 \,rad/s$.
The velocity of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$, where $x$ is the displacement.
Substituting the values: $0.2 = 2 \sqrt{0.4^2 - x^2}$.
Dividing by $2$: $0.1 = \sqrt{0.16 - x^2}$.
Squaring both sides: $0.01 = 0.16 - x^2$.
Rearranging for $x^2$: $x^2 = 0.16 - 0.01 = 0.15$.
Therefore, the displacement $x = \sqrt{0.15} \,m$.

(D) The displacement of a particle in simple harmonic motion $(SHM)$ is given by $x = a \sin(\omega t)$.
The velocity of the particle is given by $v = \omega \sqrt{a^2 - x^2}$.
Given the periodic time is $T$,the angular frequency is $\omega = \frac{2\pi}{T}$.
At $x = \frac{a}{2}$,the velocity is:
$v = \frac{2\pi}{T} \sqrt{a^2 - (\frac{a}{2})^2}$
$v = \frac{2\pi}{T} \sqrt{a^2 - \frac{a^2}{4}}$
$v = \frac{2\pi}{T} \sqrt{\frac{3a^2}{4}}$
$v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}a}{2}$
$v = \frac{\sqrt{3} \pi a}{T}$.

(C) At the mean position, the velocity is maximum, given by $v_{\max} = A\omega$.
Given $A = 4 \,cm$ and $v_{\max} = 12 \,cm/s$.
$\omega = \frac{v_{\max}}{A} = \frac{12}{4} = 3 \,rad/s$.
The velocity $v$ at any displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.
Squaring both sides: $v^2 = \omega^2 (A^2 - x^2)$.
Rearranging for $x$: $x^2 = A^2 - \frac{v^2}{\omega^2}$.
Substituting the values $A = 4$, $v = 6$, and $\omega = 3$:
$x^2 = 4^2 - \frac{6^2}{3^2} = 16 - \frac{36}{9} = 16 - 4 = 12$.
$x = \sqrt{12} = 2\sqrt{3} \,cm$.

(C) For a particle in Simple Harmonic Motion $(SHM)$,the velocity $V$ at a displacement $x$ is given by the formula: $V = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is the angular frequency. Squaring both sides,we get $V^2 = \omega^2(A^2 - x^2)$.
Applying this to the given conditions:
$V_1^2 = \omega^2(A^2 - x_1^2)$ --- $(1)$
$V_2^2 = \omega^2(A^2 - x_2^2)$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{V_1^2}{V_2^2} = \frac{A^2 - x_1^2}{A^2 - x_2^2}$
Cross-multiplying gives:
$V_1^2(A^2 - x_2^2) = V_2^2(A^2 - x_1^2)$
$V_1^2 A^2 - V_1^2 x_2^2 = V_2^2 A^2 - V_2^2 x_1^2$
Rearranging terms to solve for $A^2$:
$A^2(V_1^2 - V_2^2) = V_1^2 x_2^2 - V_2^2 x_1^2$
$A^2 = \frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}$
Taking the square root,we find the amplitude:
$A = \sqrt{\frac{V_1^2 x_2^2 - V_2^2 x_1^2}{V_1^2 - V_2^2}}$

(C) Given:
- The initial maximum velocity is $V = A\omega$.
- The amplitude doubles: $A' = 2A$.
- The period becomes one-third: $T' = \frac{T}{3}$.
Step $1$: Calculate the new angular frequency $\omega'$.
Since $\omega = \frac{2\pi}{T}$,the new angular frequency is:
$\omega' = \frac{2\pi}{T'} = \frac{2\pi}{T/3} = 3 \left(\frac{2\pi}{T}\right) = 3\omega$.
Step $2$: Calculate the new maximum velocity $V'$.
The formula for maximum velocity is $V_{\max} = A\omega$.
$V' = A' \omega' = (2A)(3\omega) = 6(A\omega)$.
Since $V = A\omega$,we get:
$V' = 6V$.

(D) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.
The acceleration in $SHM$ is given by $a = -\omega^2 x$. Considering magnitudes,$a_1 = \omega^2 |x_1|$ and $a_2 = \omega^2 |x_2|$.
The velocity in $SHM$ is given by $v^2 = \omega^2 (A^2 - x^2)$.
For the first instant: $u^2 = \omega^2 A^2 - \omega^2 x_1^2$ . . . $(i)$
For the second instant: $V^2 = \omega^2 A^2 - \omega^2 x_2^2$ . . . $(ii)$
Subtracting $(ii)$ from $(i)$: $u^2 - V^2 = \omega^2 (x_2^2 - x_1^2) = \omega^2 (x_2 - x_1)(x_2 + x_1)$ . . . $(iii)$
From the acceleration equations: $a_2 - a_1 = \omega^2 (x_2 - x_1)$ (assuming $x_2 > x_1$).
However,the distance between positions is $|x_2 - x_1|$.
Using $a_1 = \omega^2 x_1$ and $a_2 = \omega^2 x_2$,we have $x_1 = a_1/\omega^2$ and $x_2 = a_2/\omega^2$.
Substituting into the velocity difference: $u^2 - V^2 = \omega^2 (x_2^2 - x_1^2) = \omega^2 (a_2^2/\omega^4 - a_1^2/\omega^4) = \frac{1}{\omega^2} (a_2^2 - a_1^2)$.
This implies $\omega^2 = \frac{a_2^2 - a_1^2}{u^2 - V^2}$.
The distance $d = |x_2 - x_1| = |\frac{a_2 - a_1}{\omega^2}| = |\frac{a_2 - a_1}{(a_2^2 - a_1^2)/(u^2 - V^2)}| = |\frac{u^2 - V^2}{a_2 + a_1}|$.
Thus,the distance is $\frac{u^2 - V^2}{a_1 + a_2}$.

(B) The speed $V$ of a particle performing $S.H.M.$ at a displacement $x$ from the mean position is given by the formula: $V = \omega \sqrt{a^2 - x^2}$.
Given,angular frequency $\omega = \frac{2 \pi}{T}$ and displacement $x = \frac{a}{3}$.
Substituting these values into the formula:
$V = \frac{2 \pi}{T} \sqrt{a^2 - (\frac{a}{3})^2}$
$V = \frac{2 \pi}{T} \sqrt{a^2 - \frac{a^2}{9}}$
$V = \frac{2 \pi}{T} \sqrt{\frac{8 a^2}{9}}$
$V = \frac{2 \pi}{T} \times \frac{2 \sqrt{2} a}{3}$
$V = \frac{4 \sqrt{2} \pi a}{3 T}$

(C) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.
Maximum velocity is $V_{\max} = \omega A$.
Given that the speed $V = \frac{V_{\max}}{3} = \frac{\omega A}{3}$.
Equating the two expressions: $\frac{\omega A}{3} = \omega \sqrt{A^2 - x^2}$.
Squaring both sides: $\frac{A^2}{9} = A^2 - x^2$.
Rearranging for $x^2$: $x^2 = A^2 - \frac{A^2}{9} = \frac{8A^2}{9}$.
Taking the square root: $x = \sqrt{\frac{8}{9} A^2} = \frac{2\sqrt{2}}{3} A$.

(A) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.
In $S.H.M.$,acceleration $a = -\omega^2 x$. Considering magnitudes,$\alpha = \omega^2 |x_1|$ and $\beta = \omega^2 |x_2|$.
Thus,$|x_1| = \frac{\alpha}{\omega^2}$ and $|x_2| = \frac{\beta}{\omega^2}$.
The velocity in $S.H.M.$ is given by $v^2 = \omega^2(A^2 - x^2)$.
For the first instant: $u^2 = \omega^2 A^2 - \omega^2 x_1^2$ . . . $(i)$
For the second instant: $v^2 = \omega^2 A^2 - \omega^2 x_2^2$ . . . $(ii)$
Subtracting $(ii)$ from $(i)$: $u^2 - v^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)$.
Since $\alpha = \omega^2 x_1$ and $\beta = \omega^2 x_2$,we have $\alpha + \beta = \omega^2(x_1 + x_2)$.
Substituting this into the equation: $u^2 - v^2 = \omega^2(x_2 - x_1) \cdot \frac{\alpha + \beta}{\omega^2}$.
Therefore,the distance between the two positions is $|x_2 - x_1| = \frac{u^2 - v^2}{\alpha + \beta}$.

(D) The maximum velocity of a particle in $S.H.M.$ is given by $V = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
We know that $\omega = \frac{2\pi}{T}$,where $T$ is the periodic time.
Given that the new periodic time $T' = \frac{1}{3}T$ and the new amplitude $A' = 2A$.
The new angular frequency is $\omega' = \frac{2\pi}{T'} = \frac{2\pi}{\frac{1}{3}T} = 3 \left(\frac{2\pi}{T}\right) = 3\omega$.
The new maximum velocity $V'$ is given by $V' = A'\omega'$.
Substituting the values,$V' = (2A) \times (3\omega) = 6(A\omega) = 6V$.

(B) Given equations for displacement are:
$y_1 = 0.1 \sin \left(100 \pi t + \frac{\pi}{3} \right)$
$y_2 = 0.1 \cos (100 \pi t) = 0.1 \sin \left(100 \pi t + \frac{\pi}{2} \right)$
Velocity is the derivative of displacement with respect to time:
$v_1 = \frac{dy_1}{dt} = 0.1 \times 100 \pi \cos \left(100 \pi t + \frac{\pi}{3} \right) = 10 \pi \sin \left(100 \pi t + \frac{\pi}{3} + \frac{\pi}{2} \right) = 10 \pi \sin \left(100 \pi t + \frac{5\pi}{6} \right)$
$v_2 = \frac{dy_2}{dt} = 0.1 \times 100 \pi \cos \left(100 \pi t + \frac{\pi}{2} \right) = 10 \pi \sin \left(100 \pi t + \frac{\pi}{2} + \frac{\pi}{2} \right) = 10 \pi \sin (100 \pi t + \pi)$
The phase of $v_1$ is $\phi_1 = 100 \pi t + \frac{5\pi}{6}$ and the phase of $v_2$ is $\phi_2 = 100 \pi t + \pi$.
The phase difference $\Delta \phi = \phi_1 - \phi_2 = \frac{5\pi}{6} - \pi = -\frac{\pi}{6}$.

(B) The displacement of the particle is given by $x = a \sin (\omega t - \phi)$.
To find the velocity,we differentiate the displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} [a \sin (\omega t - \phi)] = a \omega \cos (\omega t - \phi)$.
Now,we substitute the given time $t = \frac{\phi}{\omega}$ into the velocity equation:
$v = a \omega \cos \left( \omega \left( \frac{\phi}{\omega} \right) - \phi \right)$.
$v = a \omega \cos (\phi - \phi) = a \omega \cos (0)$.
Since $\cos 0^{\circ} = 1$,we get:
$v = a \omega \times 1 = a \omega$.

(D) The correct option is $D$.
Concept: For $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
The maximum speed is $V = A \omega$.
In one complete time period $T$,the particle travels a total distance of $4A$.
The average speed is defined as the total distance divided by the total time.
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{4A}{T}$.
Since $T = \frac{2\pi}{\omega}$ and $\omega = \frac{V}{A}$,we have $T = \frac{2\pi A}{V}$.
Substituting $T$ into the average speed formula:
$\text{Average speed} = \frac{4A}{\left(\frac{2\pi A}{V}\right)} = \frac{4A \cdot V}{2\pi A} = \frac{2V}{\pi}$.

(C) The standard equation for linear $S.H.M$ is given by $x = a \sin(\omega t + \phi)$.
Comparing the given equation $x = 5 \sin(4t - \frac{\pi}{6})$ with the standard form,we get the amplitude $a = 5 \ cm$ and angular frequency $\omega = 4 \ rad/s$.
The velocity $v$ of a particle in $S.H.M$ at a displacement $x$ is given by the formula:
$v = \omega \sqrt{a^2 - x^2}$
Substituting the given values $a = 5 \ cm$,$x = 3 \ cm$,and $\omega = 4 \ rad/s$:
$v = 4 \sqrt{5^2 - 3^2}$
$v = 4 \sqrt{25 - 9}$
$v = 4 \sqrt{16}$
$v = 4 \times 4 = 16 \ cm/s$.
Therefore,the velocity of the particle is $16 \ cm/s$.

(D) In $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.
Velocity is given by $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$. At the mean position $(x = 0)$,the velocity is maximum $(v_{max} = A\omega)$.
Acceleration is given by $a = -\omega^2 x$. At the mean position $(x = 0)$,acceleration is minimum $(a = 0)$.
The restoring force $F = -kx$ is always directed towards the mean position.
The total energy $E = K.E. + P.E. = \frac{1}{2}kA^2$ remains constant at all times.
Therefore,statement $D$ is incorrect because the velocity is maximum,not minimum,at the mean position.

(D) Given,$x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.
Velocity $v = \frac{dx}{dt} = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.
The velocity is maximum when the magnitude of the sine function is maximum,i.e.,$\sin \left(\omega t + \frac{\pi}{4}\right) = -1$.
This occurs when the phase angle $\left(\omega t + \frac{\pi}{4}\right) = \frac{3\pi}{2}$.
$\omega t + \frac{\pi}{4} = \frac{3\pi}{2} \implies \omega t = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{5\pi}{4}$.
$t = \frac{5\pi}{4\omega}$.
However,checking the options provided,the question implies the time when the particle reaches the equilibrium position $(x=0)$ moving in the negative direction,where velocity magnitude is maximum. Setting $\omega t + \frac{\pi}{4} = \frac{\pi}{2}$ gives $\omega t = \frac{\pi}{4}$,so $t = \frac{\pi}{4\omega}$.

(B) Given: Period $T = \frac{2 \pi}{\sqrt{3}} \text{ s}$, total path length $= 4 \text{ cm}$.
Amplitude $A = \frac{\text{path length}}{2} = \frac{4}{2} = 2 \text{ cm}$.
Angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2 \pi / \sqrt{3}} = \sqrt{3} \text{ rad/s}$.
Magnitude of acceleration $a = \omega^2 |x|$ and magnitude of velocity $v = \omega \sqrt{A^2 - x^2}$.
Given $v = a$, so $\omega \sqrt{A^2 - x^2} = \omega^2 |x|$.
Squaring both sides: $\omega^2 (A^2 - x^2) = \omega^4 x^2$.
Dividing by $\omega^2$: $A^2 - x^2 = \omega^2 x^2$.
Substituting values: $2^2 - x^2 = (\sqrt{3})^2 x^2$.
$4 - x^2 = 3x^2 \implies 4x^2 = 4 \implies x^2 = 1$.
Thus, $x = 1 \text{ cm}$.

(A) The maximum velocity of a particle performing $S.H.M.$ is given by the formula: $V_{max} = \omega A = \frac{2 \pi}{T} A$.
Here,$A$ is the amplitude and $T$ is the periodic time.
Given that the initial maximum velocity is $v = \frac{2 \pi A}{T}$.
If the amplitude is tripled $(A' = 3A)$ and the periodic time is doubled $(T' = 2T)$,the new maximum velocity $V'_{max}$ is:
$V'_{max} = \frac{2 \pi A'}{T'} = \frac{2 \pi (3A)}{2T} = \frac{3}{2} \left( \frac{2 \pi A}{T} \right) = 1.5 v$.
Therefore,the new maximum velocity will be $1.5 v$.

(B) The acceleration $a$ of a particle in $S.H.M.$ is given by $a = \omega^2 x$,where $x$ is the displacement.
Given $a = 0.5 \,m/s^2$ and $x = 0.02 \,m$.
$\omega^2 = \frac{a}{x} = \frac{0.5}{0.02} = 25 \,rad^2/s^2$.
Taking the square root,we get $\omega = 5 \,rad/s$.
The maximum velocity $V_{\max}$ is given by $V_{\max} = A \omega$,where $A$ is the amplitude.
Given $A = 0.1 \,m$.
$V_{\max} = 0.1 \times 5 = 0.5 \,m/s$.

(D) The velocity $V$ of a particle in $S.H.M.$ at displacement $x$ is given by $V = \omega \sqrt{A^2 - x^2}$.
Maximum speed $V_{\max}$ occurs at the mean position and is given by $V_{\max} = A\omega$.
According to the problem,the speed $V$ is one-fourth of the maximum speed:
$V = \frac{1}{4} V_{\max}$
$\omega \sqrt{A^2 - x^2} = \frac{1}{4} (A\omega)$
$\sqrt{A^2 - x^2} = \frac{A}{4}$
Squaring both sides:
$A^2 - x^2 = \frac{A^2}{16}$
$x^2 = A^2 - \frac{A^2}{16} = \frac{15A^2}{16}$
$x = \frac{A\sqrt{15}}{4}$

(C) Given,$y_{1} = 2 \sin (10 t + \theta)$.
The velocity $V_{1} = \frac{dy_{1}}{dt} = 2 \times 10 \cos (10 t + \theta) = 20 \cos (10 t + \theta)$.
Given,$y_{2} = 3 \cos 10 t = 3 \sin (10 t + \frac{\pi}{2})$.
The velocity $V_{2} = \frac{dy_{2}}{dt} = 3 \times 10 \cos (10 t + \frac{\pi}{2}) = 30 \cos (10 t + \frac{\pi}{2})$.
The phase of $V_{1}$ is $\phi_{1} = 10 t + \theta$.
The phase of $V_{2}$ is $\phi_{2} = 10 t + \frac{\pi}{2}$.
The phase difference $\Delta \phi = \phi_{1} - \phi_{2} = (10 t + \theta) - (10 t + \frac{\pi}{2}) = \theta - \frac{\pi}{2}$.

(B) The velocity of a particle in $S.H.M.$ at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$.
At $x = \frac{2A}{3}$,the velocity is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\sqrt{5}}{3} \omega A$.
When the speed is tripled,the new velocity $v' = 3v = 3 \times \frac{\sqrt{5}}{3} \omega A = \sqrt{5} \omega A$.
The total energy of the $S.H.M.$ remains constant at any point,so $E' = K' + U$.
The new total energy is $E' = \frac{1}{2} m \omega^2 A'^2$.
The potential energy at $x = \frac{2A}{3}$ is $U = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (\frac{2A}{3})^2 = \frac{1}{2} m \omega^2 \frac{4A^2}{9}$.
The new kinetic energy is $K' = \frac{1}{2} m v'^2 = \frac{1}{2} m (\sqrt{5} \omega A)^2 = \frac{1}{2} m \omega^2 (5A^2)$.
Equating the total energy: $\frac{1}{2} m \omega^2 A'^2 = \frac{1}{2} m \omega^2 (\frac{4A^2}{9}) + \frac{1}{2} m \omega^2 (5A^2)$.
$A'^2 = \frac{4A^2}{9} + 5A^2 = A^2 (\frac{4+45}{9}) = \frac{49A^2}{9}$.
Taking the square root,$A' = \frac{7A}{3}$.

(A) The velocity $v$ of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega = \sqrt{\frac{k}{m}}$.
Given: $m = 10 \ kg$,$k = 10 \ N/m$,$A = 0.5 \ m$,and $v = 40 \ cm/s = 0.4 \ m/s$.
First,calculate the angular frequency $\omega = \sqrt{\frac{10}{10}} = 1 \ rad/s$.
Now,substitute the values into the velocity equation: $0.4 = 1 \cdot \sqrt{(0.5)^2 - x^2}$.
Squaring both sides: $0.16 = 0.25 - x^2$.
Rearranging for $x^2$: $x^2 = 0.25 - 0.16 = 0.09$.
Taking the square root: $x = 0.3 \ m$.

(D) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.
Given that the particle is halfway between the mean position $(x = 0)$ and the extreme position $(x = A)$,the displacement is $x = \frac{A}{2}$.
The angular frequency $\omega$ is related to the period $T$ by $\omega = \frac{2 \pi}{T}$.
Substituting these values into the velocity formula:
$v = \frac{2 \pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$
$v = \frac{2 \pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$
$v = \frac{2 \pi}{T} \sqrt{\frac{3 A^2}{4}}$
$v = \frac{2 \pi}{T} \times \frac{\sqrt{3} A}{2}$
$v = \frac{\sqrt{3} \pi A}{T}$

(D) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.
The velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
Given: $v = \pi \ m \ s^{-1}$,$T = 16 \ s$,and $t = 2 \ s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8} \ rad \ s^{-1}$.
Substituting the values into the velocity equation:
$\pi = A \times \frac{\pi}{8} \times \cos\left(\frac{\pi}{8} \times 2\right)$
$\pi = A \times \frac{\pi}{8} \times \cos\left(\frac{\pi}{4}\right)$
$1 = \frac{A}{8} \times \frac{1}{\sqrt{2}}$
$A = 8\sqrt{2} \ m$.

(B) The maximum velocity $(v_{\max})$ of a particle executing simple harmonic motion is given by the formula $v_{\max} = A \omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.
Given: Amplitude $A = 50 \,mm = 50 \times 10^{-3} \,m = 0.05 \,m$.
Time period $T = 2 \,s$.
The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2} = \pi \,rad/s$.
Substituting the values: $v_{\max} = 0.05 \times \pi \approx 0.05 \times 3.14159 = 0.157 \,ms^{-1}$.
Rounding to two decimal places, we get $v_{\max} \approx 0.16 \,ms^{-1}$. However, based on the provided options, the closest value is $0.15 \,ms^{-1}$.

(C) The velocity of a particle in $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$.
The acceleration of a particle in $SHM$ is given by $a = \omega^2 x$.
Given that the magnitude of acceleration equals the magnitude of velocity:
$\omega^2 x = \omega \sqrt{A^2 - x^2}$
$\omega x = \sqrt{A^2 - x^2}$ (Equation $i$)
Given the time period $T = \frac{2 \pi}{\sqrt{3}} \,s$,we find the angular frequency:
$\omega = \frac{2 \pi}{T} = \sqrt{3} \,rad/s$.
Substituting $\omega$ into Equation $i$:
$\sqrt{3} x = \sqrt{A^2 - x^2}$
Squaring both sides:
$3x^2 = A^2 - x^2$
$4x^2 = A^2 \Rightarrow A = 2x$.
The path length is $4 \,cm$,so the amplitude $A = \frac{\text{path length}}{2} = 2 \,cm$.
Substituting $A = 2 \,cm$ into $A = 2x$:
$2 = 2x \Rightarrow x = 1 \,cm$.

(D) Given the equation for velocity in $S.H.M.$: $4V^2 = 50 - x^2$.
Rearranging this,we get $V^2 = \frac{1}{4}(50 - x^2) = \frac{1}{4}(50 - x^2)$.
Comparing this with the standard $S.H.M.$ velocity equation $V^2 = \omega^2(A^2 - x^2)$,we can rewrite the given equation as $V^2 = \frac{1}{4} \cdot 50 - \frac{1}{4}x^2$.
This implies $\omega^2 = \frac{1}{4}$,so $\omega = \frac{1}{2} \text{ rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1/2} = 4\pi$.
Substituting $\pi = \frac{22}{7}$,we get $T = 4 \times \frac{22}{7} = \frac{88}{7} \text{ seconds}$.
Comparing this with the given time period $\frac{x}{7}$,we find $x = 88$.

(B) The velocity of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$.
Substituting the given values:
For $x_1 = 3 \ cm$,$v_1 = 8 \ cm/s$: $8 = \omega \sqrt{A^2 - 3^2} \implies 8 = \omega \sqrt{A^2 - 9}$ ... $(i)$
For $x_2 = 4 \ cm$,$v_2 = 6 \ cm/s$: $6 = \omega \sqrt{A^2 - 4^2} \implies 6 = \omega \sqrt{A^2 - 16}$ ... (ii)
Dividing equation $(i)$ by (ii):
$\frac{8}{6} = \frac{\sqrt{A^2 - 9}}{\sqrt{A^2 - 16}} \implies \frac{4}{3} = \frac{\sqrt{A^2 - 9}}{\sqrt{A^2 - 16}}$
Squaring both sides:
$\frac{16}{9} = \frac{A^2 - 9}{A^2 - 16} \implies 16(A^2 - 16) = 9(A^2 - 9)$
$16A^2 - 256 = 9A^2 - 81 \implies 7A^2 = 175 \implies A^2 = 25 \ cm^2$.
Substituting $A^2 = 25$ into equation (ii):
$6 = \omega \sqrt{25 - 16} = \omega \sqrt{9} = 3\omega \implies \omega = 2 \ rad/s$.
The periodic time $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \ s$.

(B) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.
Velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t) \dots (i)$.
Given: $v = \pi \ m/s$,$T = 12 \ s$,and $t = 2 \ s$.
Angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{12} = \frac{\pi}{6} \ rad/s$.
Substituting these values into equation $(i)$:
$\pi = A \times \frac{\pi}{6} \times \cos\left(\frac{\pi}{6} \times 2\right)$
$\pi = A \times \frac{\pi}{6} \times \cos\left(\frac{\pi}{3}\right)$
Since $\cos(60^{\circ}) = 0.5 = \frac{1}{2}$,we have:
$1 = \frac{A}{6} \times \frac{1}{2}$
$1 = \frac{A}{12}$
$A = 12 \ m$.

(D) The maximum velocity $(V_{\max})$ of a particle in Simple Harmonic Motion ($S$.$H$.$M$.) is given by the formula: $V_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given: Amplitude $A = 7 \ mm = 7 \times 10^{-3} \ m$,Maximum velocity $V_{\max} = 4.4 \ ms^{-1}$,and $\pi = \frac{22}{7}$.
We know that $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
Substituting the values into the formula: $4.4 = (7 \times 10^{-3}) \times \left(\frac{2 \times 22/7}{T}\right)$.
Simplifying the expression: $4.4 = (7 \times 10^{-3}) \times \left(\frac{44}{7T}\right)$.
$4.4 = \frac{44 \times 10^{-3}}{T}$.
$T = \frac{44 \times 10^{-3}}{4.4} = 10 \times 10^{-3} = 0.01 \ s$.

(C) The velocity of a particle in $S.H.M.$ at a distance $x$ from the mean position is given by $V = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Squaring both sides,we get $V^2 = \omega^2(A^2 - x^2)$.
For the given conditions:
$V_1^2 = \omega^2(A^2 - x_1^2)$ $(i)$
$V_2^2 = \omega^2(A^2 - x_2^2)$ $(ii)$
Subtracting equation $(i)$ from $(ii)$:
$V_2^2 - V_1^2 = \omega^2(A^2 - x_2^2 - A^2 + x_1^2)$
$V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2)$
$\omega^2 = \frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}$
$\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
Since the frequency $f = \frac{\omega}{2\pi}$,we have:
$f = \frac{1}{2\pi} \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$

(C) For a particle executing $SHM$,the displacement is $x = A \sin(\omega t + \phi)$.
Velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
Acceleration is $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = -\omega^2 x$.
We know that angular frequency $\omega = \frac{2 \pi}{T}$.
Now,let us evaluate the expression in option $(c)$:
$\frac{a T}{v} = \frac{(-\omega^2 x) T}{A \omega \cos(\omega t + \phi)} = \frac{-\omega^2 (A \sin(\omega t + \phi)) T}{A \omega \cos(\omega t + \phi)} = -\omega T \tan(\omega t + \phi)$.
Wait,let's re-evaluate using the standard relation $a = -\omega^2 x$ and $v = \omega \sqrt{A^2 - x^2}$.
Actually,the ratio $\frac{a}{v}$ for $SHM$ is $\frac{-\omega^2 x}{v}$. Since $v = \omega \sqrt{A^2 - x^2}$,$\frac{a}{v} = \frac{-\omega^2 x}{\omega \sqrt{A^2 - x^2}} = -\omega \frac{x}{\sqrt{A^2 - x^2}}$. This is time-dependent.
However,checking the dimensions: $[a] = [L T^{-2}]$,$[T] = [T]$,$[v] = [L T^{-1}]$.
Dimension of $\frac{a T}{v} = \frac{[L T^{-2}] [T]}{[L T^{-1}]} = \frac{[L T^{-1}]}{[L T^{-1}]} = [M^0 L^0 T^0]$.
Since the expression $\frac{a T}{v}$ is dimensionless and represents a constant factor related to the angular frequency $(2\pi)$,it is the only quantity that remains constant in magnitude relative to the phase of motion.

(B) We know that for a particle executing $SHM$,the velocity $v$ and acceleration $a$ are given by the equations:
$v = \omega \sqrt{A^2 - y^2}$
$a = -\omega^2 y$
At the mean position,the displacement $y = 0$.
Substituting $y = 0$ into the velocity equation:
$v = \omega \sqrt{A^2 - 0} = \omega A$
Thus,the velocity is maximum $(v_{\max} = \omega A)$.
Substituting $y = 0$ into the acceleration equation:
$a = -\omega^2 (0) = 0$
Thus,the acceleration is zero.

(B) The displacement equation for a particle executing $SHM$ is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$ with the standard form:
Amplitude $A = 3 \ m$.
Angular frequency $\omega = 2 \pi \ rad/s$.
The maximum speed $v_{\max}$ of a particle in $SHM$ is given by the formula $v_{\max} = \omega A$.
Substituting the values: $v_{\max} = (2 \pi \ rad/s) \times (3 \ m) = 6 \pi \ m/s$.
Thus,the amplitude is $3 \ m$ and the maximum speed is $6 \pi \ m/s$.

(D) The position of the particle is given by $x = 2 \cos (2t)$.
Taking the derivative with respect to time $t$,we get the velocity $v$:
$v = \frac{dx}{dt} = \frac{d}{dt} [2 \cos (2t)] = -2 \sin (2t) \times 2 = -4 \sin (2t) \,m/s$.
The maximum velocity $v_{\max}$ occurs when $|\sin (2t)| = 1$,so $|v_{\max}| = 4 \,m/s$.
The maximum kinetic energy $K_{\max}$ is given by the formula $K_{\max} = \frac{1}{2} m (v_{\max})^2$.
Substituting the given mass $m = 2 \,kg$ and $v_{\max} = 4 \,m/s$:
$K_{\max} = \frac{1}{2} \times 2 \,kg \times (4 \,m/s)^2 = 16 \,J$.

(A) For a particle executing simple harmonic motion $(SHM)$, the maximum velocity is given by $v_{max} = A\omega$ and the maximum acceleration is given by $a_{max} = A\omega^2$, where $A$ is the amplitude and $\omega$ is the angular frequency.
Given: $v_{max} = 5 \,m \,s^{-1}$ and $a_{max} = 10 \,m \,s^{-2}$.
Dividing the expression for $a_{max}$ by $v_{max}$, we get: $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$.
Therefore, $\omega = \frac{10}{5} = 2 \,rad \,s^{-1}$.
The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$, we get $T = \frac{2\pi}{2} = \pi \,s$.

(D) In simple harmonic motion,the velocity $v$ at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Initially,the velocity is $v_1 = \omega \sqrt{A^2 - x^2}$.
After the blow,the velocity becomes $v_2 = 2v_1 = 2\omega \sqrt{A^2 - x^2}$.
Let the new amplitude be $A'$. Since the frequency $\omega$ remains constant,the new velocity at the same position $x$ is $v_2 = \omega \sqrt{A'^2 - x^2}$.
Equating the two expressions for $v_2$: $\omega \sqrt{A'^2 - x^2} = 2\omega \sqrt{A^2 - x^2}$.
Squaring both sides: $A'^2 - x^2 = 4(A^2 - x^2)$.
$A'^2 - x^2 = 4A^2 - 4x^2$.
$A'^2 = 4A^2 - 3x^2$.
Therefore,the new amplitude is $A' = \sqrt{4A^2 - 3x^2}$.

(C) The displacement of the first particle is $y_1 = 0.1 \sin(100 \pi t + \frac{\pi}{3})$. The velocity $v_1$ is given by $v_1 = \frac{dy_1}{dt} = 0.1 \times 100 \pi \cos(100 \pi t + \frac{\pi}{3}) = 10 \pi \sin(100 \pi t + \frac{\pi}{3} + \frac{\pi}{2})$.
At $t=0$,the phase of $v_1$ is $\phi_1 = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}$.
The displacement of the second particle is $y_2 = 0.1 \cos(\pi t) = 0.1 \sin(\pi t + \frac{\pi}{2})$. The velocity $v_2$ is given by $v_2 = \frac{dy_2}{dt} = 0.1 \times \pi \cos(\pi t + \frac{\pi}{2}) = 0.1 \pi \sin(\pi t + \frac{\pi}{2} + \frac{\pi}{2}) = 0.1 \pi \sin(\pi t + \pi)$.
At $t=0$,the phase of $v_2$ is $\phi_2 = \pi$.
The phase difference is $\Delta \phi = |\phi_2 - \phi_1| = |\pi - \frac{5\pi}{6}| = \frac{\pi}{6}$.

(A) The displacement of a particle in $SHM$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.
Given $A = 10 \,cm$ and $x = 5 \,cm$,we have $5 = 10 \sin(\omega t)$,which implies $\sin(\omega t) = 1/2$.
Thus,$\omega t = \pi/6$. Since $\omega = 2\pi/T$,we get $t = \frac{\pi/6}{2\pi/T} = T/12$.
Given $T = 0.6 \,s$,the time interval is $t = 0.6/12 = 0.05 \,s$.
The mean velocity $V_{\text{mean}}$ is defined as the total displacement divided by the total time interval:
$V_{\text{mean}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$.
Starting from equilibrium ($x_i = 0$ at $t_i = 0$) to $x_f = 5 \,cm$ at $t_f = 0.05 \,s$:
$V_{\text{mean}} = \frac{5 \,cm - 0}{0.05 \,s} = 100 \,cm/s = 1 \,m/s$.

(A) Given, mass $m = 400 \, g = 0.4 \, kg$.
Force $F = -10 x$.
Comparing with the standard equation of $S.H.M.$, $F = -m \omega^2 x$, we get $m \omega^2 = 10$.
Substituting the value of $m$:
$0.4 \omega^2 = 10$
$\omega^2 = \frac{10}{0.4} = 25$
$\omega = 5 \, rad/s$.
The maximum speed at the centre of oscillation is given by $V_{max} = \omega A$.
Given $V_{max} = 10 \, m/s$, we have:
$10 = 5 \times A$
$A = \frac{10}{5} = 2 \, m$.