Velocity of Simple Harmonic Motion Questions in English

(C) The maximum speed of a particle in $S.H.M.$ is given by $v_{\max} = \omega A$.
The speed $v$ at any displacement $y$ is given by $v = \omega \sqrt{A^2 - y^2}$.
According to the problem,the speed is half of the maximum speed,so $v = \frac{v_{\max}}{2} = \frac{\omega A}{2}$.
Equating the two expressions for speed:
$\frac{\omega A}{2} = \omega \sqrt{A^2 - y^2}$
Squaring both sides:
$\frac{A^2}{4} = A^2 - y^2$
Rearranging to solve for $y^2$:
$y^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$
Taking the square root:
$y = \frac{\sqrt{3}A}{2}$

(A) The maximum speed $(v_{\max})$ of a particle in simple harmonic motion is given by the formula $v_{\max} = A\omega$,where $A$ is the amplitude (maximum displacement) and $\omega$ is the angular frequency.
Given that $\omega = \frac{2\pi}{T}$,we can write $v_{\max} = A \times \frac{2\pi}{T}$.
Rearranging for $A$,we get $A = \frac{v_{\max} \times T}{2\pi}$.
Substituting the given values: $v_{\max} = 1.00 \times 10^3 \, m/s$ and $T = 1.00 \times 10^{-5} \, s$.
$A = \frac{(1.00 \times 10^3) \times (1.00 \times 10^{-5})}{2\pi} = \frac{1.00 \times 10^{-2}}{2\pi} \, m$.
$A = \frac{0.01}{6.283} \approx 0.00159 \, m = 1.59 \, mm$.

(A) The velocity $v$ of a particle executing simple harmonic motion $(S.H.M.)$ at a displacement $x$ is given by the formula:
$v = \omega \sqrt{A^2 - x^2}$
Given that the angular frequency $\omega = \frac{2\pi}{T}$ and the displacement $x = \frac{A}{2}$,we substitute these values into the equation:
$v = \frac{2\pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$
Simplify the expression inside the square root:
$v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$
$v = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}}$
Extracting the terms from the square root:
$v = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2}$
$v = \frac{\pi A\sqrt{3}}{T}$
Thus,the speed of the pendulum at $X = \frac{A}{2}$ is $\frac{\pi A\sqrt{3}}{T}$.

(C) The velocity $v$ of a particle executing simple harmonic motion at a displacement $y$ is given by the formula: $v = \omega \sqrt{a^2 - y^2}$.
Given:
Angular frequency $\omega = 2\,rad/s$
Amplitude $a = 60\,mm$
Displacement $y = 20\,mm$
Substituting these values into the formula:
$v = 2 \sqrt{60^2 - 20^2}$
$v = 2 \sqrt{3600 - 400}$
$v = 2 \sqrt{3200}$
$v = 2 \times 56.568$
$v \approx 113.14\,mm/s$.
Rounding to the nearest integer,we get $113\,mm/s$.

(C) Given: Amplitude $a = 10\, cm$,Maximum velocity $v_{\max} = 100\, cm/s$.
We know that $v_{\max} = a\omega$.
Therefore,$\omega = \frac{v_{\max}}{a} = \frac{100}{10} = 10\, rad/s$.
The velocity $v$ at a displacement $y$ is given by the formula $v = \omega \sqrt{a^2 - y^2}$.
Substituting the given values: $50 = 10 \sqrt{10^2 - y^2}$.
Dividing by $10$: $5 = \sqrt{100 - y^2}$.
Squaring both sides: $25 = 100 - y^2$.
$y^2 = 100 - 25 = 75$.
$y = \sqrt{75} = 5\sqrt{3}\, cm$.

(C) The velocity of a simple harmonic oscillator is given by $v = \omega \sqrt{A^2 - x^2}$.
At the centre of oscillation,the displacement $x = 0$.
Therefore,the maximum velocity is $v_{\text{max}} = A\omega$.
Given,amplitude $A = 0.2 \, m$ and time period $T = 0.01 \, s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.01} = 200\pi \, rad/s$.
Substituting these values,we get $v_{\text{max}} = 0.2 \times 200\pi = 40\pi \, m/s$.

(B) The maximum speed of a particle executing $S.H.M.$ is given by the formula $v_{\max} = a\omega$.
Here,$a$ is the amplitude and $\omega$ is the angular frequency.
The angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$.
Given: Amplitude $a = 3 \, cm$ and time period $T = 6 \, s$.
Substituting these values into the formula:
$v_{\max} = a \times \frac{2\pi}{T} = 3 \times \frac{2\pi}{6} = \pi \, cm/s$.
Therefore,the correct option is $B$.

(C) The maximum velocity $(v_{\max})$ of a particle executing $S.H.M.$ is given by the formula: $v_{\max} = \omega a$.
Here,$\omega$ is the angular frequency,which is defined as $\omega = \frac{2\pi}{T}$.
Given: Amplitude $a = 2 \, m$ and periodic time $T = 2 \, s$.
Substituting these values into the formula:
$v_{\max} = \left( \frac{2\pi}{T} \right) \times a$
$v_{\max} = \left( \frac{2\pi}{2} \right) \times 2$
$v_{\max} = \pi \times 2 = 2\pi \, m/s$.
Therefore,the maximum velocity is $2\pi \, m/s$.

(D) The displacement of a particle in $S.H.M.$ is given by $x = a \sin(\omega t + \phi)$.
Taking the derivative with respect to time,the velocity is $v = \frac{dx}{dt} = a\omega \cos(\omega t + \phi)$.
The maximum velocity occurs when $\cos(\omega t + \phi) = 1$,so $v_{\max} = a\omega$.
Since the angular frequency $\omega = \frac{2\pi}{T}$,substituting this into the expression gives $v_{\max} = a \left(\frac{2\pi}{T}\right) = \frac{2\pi a}{T}$.

(C) The velocity $v$ of a body executing $S.H.M.$ at a displacement $y$ is given by the formula: $v = \omega \sqrt{a^2 - y^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.
For $y_1 = 4 \, cm$,$v_1 = 10 \, cm/sec$: $10 = \omega \sqrt{a^2 - 4^2} \implies 100 = \omega^2(a^2 - 16)$ --- $(1)$
For $y_2 = 5 \, cm$,$v_2 = 8 \, cm/sec$: $8 = \omega \sqrt{a^2 - 5^2} \implies 64 = \omega^2(a^2 - 25)$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$100 - 64 = \omega^2(a^2 - 16 - a^2 + 25)$
$36 = \omega^2(9)$
$\omega^2 = 4 \implies \omega = 2 \, rad/sec$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi \, sec$.

(D) The equation of motion is given by $x = 5 \sin \left( 4t - \frac{\pi}{6} \right)$.
Comparing this with the standard equation $x = A \sin(\omega t + \phi)$,we get the amplitude $A = 5$ and the angular frequency $\omega = 4$.
The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by the formula $v = \omega \sqrt{A^2 - x^2}$.
Substituting the given values $A = 5$,$\omega = 4$,and $x = 3$:
$v = 4 \sqrt{5^2 - 3^2}$
$v = 4 \sqrt{25 - 9}$
$v = 4 \sqrt{16}$
$v = 4 \times 4 = 16$ units.

(B) The maximum velocity $(v_{\max})$ of a simple harmonic oscillator 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: $v_{\max} = 0.05 \times \pi$.
Using $\pi \approx 3.14159$,we get $v_{\max} = 0.05 \times 3.14159 \approx 0.157\, m/s$.
Rounding to two decimal places,we get $0.16\, m/s$.
Thus,the correct option is $B$.

(D) For a body in simple harmonic motion,the maximum velocity is given by $v_{\max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular velocity.
The maximum acceleration is given by $a_{\max} = A\omega^2$.
Dividing the expression for maximum acceleration by the expression for maximum velocity:
$\frac{a_{\max}}{v_{\max}} = \frac{A\omega^2}{A\omega} = \omega$.
Given $v_{\max} = 2 \ m/s$ and $a_{\max} = 4 \ m/s^2$,we have:
$\omega = \frac{4}{2} = 2 \ rad/s$.

(A) The maximum velocity $(v_{\max})$ of a particle performing $S.H.M.$ is given by the formula $v_{\max} = a\omega$.
Here,$a$ is the amplitude and $\omega$ is the angular frequency.
The angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$.
Given: Amplitude $a = 2 \times 10^{-3} \ m$ and Time period $T = 0.1 \ s$.
Substituting these values into the formula:
$v_{\max} = a \times \frac{2\pi}{T} = (2 \times 10^{-3}) \times \frac{2\pi}{0.1}$.
$v_{\max} = \frac{4 \times 10^{-3} \times \pi}{10^{-1}} = 4 \times 10^{-2} \times \pi = 0.04\pi \ m/s$.
$v_{\max} = \frac{4\pi}{100} = \frac{\pi}{25} \ m/s$.
Therefore,the correct option is $A$.

(C) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Given $v_{\max} = 10 \, cm/s$ and $a = 4 \, cm$,we have $\omega = \frac{v_{\max}}{a} = \frac{10}{4} = 2.5 \, rad/s$.
The velocity $v$ at a displacement $y$ from the mean position is given by $v = \omega \sqrt{a^2 - y^2}$.
Squaring both sides,we get $v^2 = \omega^2(a^2 - y^2)$.
Rearranging for $y$,we get $y^2 = a^2 - \frac{v^2}{\omega^2}$,so $y = \sqrt{a^2 - \frac{v^2}{\omega^2}}$.
Substituting the values $a = 4 \, cm$,$v = 5 \, cm/s$,and $\omega = 2.5 \, rad/s$:
$y = \sqrt{4^2 - \frac{5^2}{(2.5)^2}} = \sqrt{16 - \frac{25}{6.25}} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \, cm$.

(B) For a particle performing simple harmonic motion $(SHM)$ with amplitude $A$ and angular velocity $\omega$:
Maximum velocity is given by $v_{\max} = A\omega$.
Maximum acceleration is given by $a_{\max} = A\omega^2$.
The ratio of maximum velocity to maximum acceleration is $\frac{v_{\max}}{a_{\max}} = \frac{A\omega}{A\omega^2} = \frac{1}{\omega}$.

(A) 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 velocity.
Given that all three bodies have the same maximum velocity,we have:
$v_1 = v_2 = v_3$
Substituting the formula for maximum velocity for each body:
$A_1\omega_1 = A_2\omega_2 = A_3\omega_3$
Therefore,the correct relation is ${A_1}{\omega _1} = {A_2}{\omega _2} = {A_3}{\omega _3}$.

(D) In Simple Harmonic Motion $(S.H.M.)$,the displacement $x$ is given by $x = A \sin(\omega t + \phi)$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
At the mean position,the displacement $x = 0$,which implies $\sin(\omega t + \phi) = 0$,so $\cos(\omega t + \phi) = \pm 1$.
Therefore,the magnitude of velocity is $|v| = A\omega$,which is the maximum possible value for the velocity in $S.H.M.$

(B) The displacement of a particle in simple harmonic motion is given by $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Differentiating the displacement with respect to time $t$ gives the velocity $v$:
$v = \frac{dy}{dt} = A \omega \cos(\omega t)$
Using the trigonometric identity $\cos(\omega t) = \sqrt{1 - \sin^2(\omega t)}$,we substitute $\sin(\omega t) = \frac{y}{A}$:
$v = A \omega \sqrt{1 - (\frac{y}{A})^2}$
Simplifying the expression:
$v = A \omega \sqrt{\frac{A^2 - y^2}{A^2}}$
$v = A \omega \frac{\sqrt{A^2 - y^2}}{A}$
$v = \omega \sqrt{A^2 - y^2}$
Thus,the velocity of the particle at displacement $y$ is $\omega \sqrt{A^2 - y^2}$.

(B) The displacement of the particle is given by $x = A \cos (\omega t - \theta)$.
To find the velocity $v$,we differentiate the displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} [A \cos (\omega t - \theta)]$
$v = -A \omega \sin (\omega t - \theta)$
The velocity is maximum when the magnitude of the sine function is $1$,i.e.,$|\sin (\omega t - \theta)| = 1$.
Therefore,the maximum velocity is $v_{\max} = | -A \omega (1) | = A \omega$.
Thus,the maximum velocity of the particle is $A \omega$.

(D) For a particle executing $SHM$,the velocity $v$ at a displacement $y$ from the mean position is given by $v = \omega \sqrt{A^2 - y^2}$,where $A$ is the amplitude.
At the mean position $(y = 0)$,the velocity is maximum,given by $v_{max} = A\omega$.
Given $A = 4 \,cm$ and $v_{max} = 16 \,cm/s$,we have $16 = 4\omega$,which implies $\omega = 4 \,rad/s$.
Now,we need to find the distance $y$ where $v = 8\sqrt{3} \,cm/s$.
Using the formula $v = \omega \sqrt{A^2 - y^2}$:
$8\sqrt{3} = 4 \sqrt{4^2 - y^2}$
Divide both sides by $4$:
$2\sqrt{3} = \sqrt{16 - y^2}$
Square both sides:
$(2\sqrt{3})^2 = 16 - y^2$
$4 \times 3 = 16 - y^2$
$12 = 16 - y^2$
$y^2 = 16 - 12 = 4$
$y = 2 \,cm$.

(A) The given equation for simple harmonic motion is $y = a \sin(\omega t + \phi)$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Comparing this with the given equation $y = 3 \sin \left( 100t + \frac{\pi}{6} \right)$,we get:
Amplitude $a = 3$
Angular frequency $\omega = 100 \text{ rad/s}$.
The maximum velocity $v_{\max}$ of a particle in simple harmonic motion is given by the formula $v_{\max} = a\omega$.
Substituting the values,we get $v_{\max} = 3 \times 100 = 300$.

(A) Given: Amplitude $A = 0.06\, m$,Frequency $n = 15\, Hz$.
Angular frequency $\omega = 2\pi n = 2 \times 3.14159 \times 15 = 94.248\, rad/s$.
Maximum velocity $v_{max} = A\omega = 0.06 \times 94.248 = 5.654\, m/s$.
Maximum acceleration $a_{max} = A\omega^2 = 0.06 \times (94.248)^2 = 0.06 \times 8882.68 = 532.96\, m/s^2 = 5.33 \times 10^2\, m/s^2$.
Thus,the correct option is $A$.

(C) The kinetic energy of a particle in $S.H.M.$ is maximum at the mean position.
Given,maximum kinetic energy,$K_{\max} = 16 \, J$.
The mass of the particle,$m = 0.32 \, kg$.
The formula for maximum kinetic energy is $K_{\max} = \frac{1}{2} m v_{\max}^2$.
Rearranging for maximum velocity,$v_{\max} = \sqrt{\frac{2 K_{\max}}{m}}$.
Substituting the values,$v_{\max} = \sqrt{\frac{2 \times 16}{0.32}}$.
$v_{\max} = \sqrt{\frac{32}{0.32}} = \sqrt{100}$.
$v_{\max} = 10 \, m/s$.

(D) Given: Maximum velocity $v_{max} = \omega A = 1 \ m/s$ and maximum acceleration $a_{max} = \omega^2 A = 1.57 \ m/s^2$.
Dividing the expression for maximum acceleration by the expression for maximum velocity:
$\frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} = \omega$
$\omega = \frac{1.57}{1} = 1.57 \ rad/s$.
We know that $\omega = \frac{2\pi}{T}$.
Substituting $\pi \approx 3.14$:
$1.57 = \frac{2 \times 3.14}{T}$
$T = \frac{6.28}{1.57} = 4 \ s$.
Therefore,the time period of the particle is $4 \ s$.

(B) Given,initial position $x = 1 \, cm$,initial velocity $v = \pi \, cm/s$,and angular frequency $\omega = \pi \, s^{-1}$.
Using the relation between velocity and displacement in $SHM$:
$v = \omega \sqrt{A^2 - x^2}$
Substituting the given values:
$\pi = \pi \sqrt{A^2 - (1)^2}$
Dividing both sides by $\pi$:
$1 = \sqrt{A^2 - 1}$
Squaring both sides:
$1 = A^2 - 1$
$A^2 = 2$
$A = \sqrt{2} \, cm$.
Thus,the amplitude is $\sqrt{2} \, cm$.

(B) Given: Mass $m = 10\,kg$,Force constant $k = 10\,N/m$,Velocity $v = 40\,cm/s = 0.4\,m/s$,Amplitude $A = 0.5\,m$.
First,calculate the angular frequency $\omega$:
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{10}} = 1\,rad/s$.
The velocity $v$ at displacement $y$ is given by the formula:
$v = \omega \sqrt{A^2 - y^2}$.
Substituting the values:
$0.4 = 1 \times \sqrt{(0.5)^2 - y^2}$.
Squaring both sides:
$(0.4)^2 = (0.5)^2 - y^2$.
$0.16 = 0.25 - y^2$.
Solving for $y^2$:
$y^2 = 0.25 - 0.16 = 0.09$.
Taking the square root:
$y = \sqrt{0.09} = 0.3\,m$.

(A) From the given velocity-time graph,we can observe that one complete cycle of the oscillation is completed in a time interval of $T = 0.04 \; s$.
The frequency of oscillation $(f)$ is defined as the reciprocal of the time period $(T)$:
$f = \frac{1}{T}$
Substituting the value of $T$ from the graph:
$f = \frac{1}{0.04} \; Hz$
$f = 25 \; Hz$
Therefore,the frequency of oscillation is $25 \; Hz$.

(D) In simple harmonic motion $(S.H.M.)$,the displacement $y$ is given by $y = a \sin(\omega t)$ and velocity $v$ is given by $v = a \omega \cos(\omega t)$.
From these equations,we can write $\sin(\omega t) = \frac{y}{a}$ and $\cos(\omega t) = \frac{v}{a \omega}$.
Using the trigonometric identity $\sin^2(\omega t) + \cos^2(\omega t) = 1$,we get:
$\frac{y^2}{a^2} + \frac{v^2}{a^2 \omega^2} = 1$.
This is the standard equation of an ellipse in the $(y, v)$ plane,where $y$ is the displacement and $v$ is the velocity.

(C) The velocity $v$ of a particle in simple harmonic motion at displacement $y$ is given by $v = \omega \sqrt{A^2 - y^2}$.
The maximum velocity is $v_{\text{max}} = A\omega$.
Given that the velocity is half of the maximum velocity,we have $v = \frac{A\omega}{2}$.
Substituting this into the velocity equation:
$\frac{A\omega}{2} = \omega \sqrt{A^2 - y^2}$
Squaring both sides:
$\frac{A^2}{4} = A^2 - y^2$
Rearranging for $y^2$:
$y^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$
Taking the square root:
$y = \frac{\sqrt{3}A}{2}$.

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

(A) 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 the angular frequency $\omega = \frac{2\pi}{T}$ and the displacement $x = \frac{A}{2}$,we substitute these values into the formula:
$v = \frac{2\pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$
$v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$
$v = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}}$
$v = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2}$
$v = \frac{\pi A \sqrt{3}}{T}$.

(C) The maximum velocity $(v_{\max})$ of a particle in 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 $a = 2 \ m$
Time period $T = 2 \ s$
We know that $\omega = \frac{2\pi}{T}$.
Substituting the values:
$\omega = \frac{2\pi}{2} = \pi \ rad/s$
Now,calculating $v_{\max}$:
$v_{\max} = 2 \times \pi = 2\pi \ m/s$
Therefore,the correct option is $C$.

(D) The displacement of a particle in simple harmonic motion is given by $x = a \sin(\omega t + \phi)$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = a\omega \cos(\omega t + \phi)$.
The maximum velocity $V_{max}$ occurs when $\cos(\omega t + \phi) = 1$,so $V_{max} = a\omega$.
Since the angular frequency $\omega = \frac{2\pi}{T}$,substituting this into the expression gives $V_{max} = a \left(\frac{2\pi}{T}\right) = \frac{2\pi a}{T}$.

(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 $a = 3 \ cm$,Time period $T = 6 \ s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \ rad/s$.
Substituting these values into the formula:
$v_{\max} = 3 \times \frac{\pi}{3} = \pi \ cm/s$.
Therefore,the maximum velocity is $\pi \ cm/s$.

(C) Given: Amplitude $a = 10 \ cm$,Maximum velocity $v_{\max} = 100 \ cm/s$.
We know that $v_{\max} = a\omega$,so $100 = 10 \times \omega$,which gives $\omega = 10 \ rad/s$.
The velocity $v$ at a displacement $y$ is given by the formula $v = \omega \sqrt{a^2 - y^2}$.
Substituting the given values: $50 = 10 \sqrt{10^2 - y^2}$.
Dividing by $10$: $5 = \sqrt{100 - y^2}$.
Squaring both sides: $25 = 100 - y^2$.
$y^2 = 100 - 25 = 75$.
$y = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \ cm$.

(A) For a particle in simple harmonic motion,the maximum acceleration is given by $A_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Given $A_{max} = 24 \ m/s^2$,so $\omega^2 A = 24$ --- $(i)$
The maximum velocity is given by $V_{max} = \omega A$.
Given $V_{max} = 16 \ m/s$,so $\omega A = 16$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{\omega^2 A}{\omega A} = \frac{24}{16}$
$\omega = 1.5 \ rad/s = 3/2 \ rad/s$
Substitute $\omega = 3/2$ into equation $(ii)$:
$(3/2) A = 16$
$A = 16 \times (2/3) = 32/3 \ m$
Therefore,the amplitude is $32/3 \ m$.

(C) For simple harmonic motion,the velocity $v$ at position $x$ is given by $v = \omega \sqrt{a^2 - x^2}$.
Given the amplitude is $a$ and the time period is $T$,the angular frequency is $\omega = \frac{2\pi}{T}$.
At position $x = \frac{a}{2}$,the speed is:
$v = \omega \sqrt{a^2 - (\frac{a}{2})^2}$
$v = \omega \sqrt{a^2 - \frac{a^2}{4}}$
$v = \omega \sqrt{\frac{3a^2}{4}}$
$v = \omega \cdot \frac{a\sqrt{3}}{2}$
Substituting $\omega = \frac{2\pi}{T}$:
$v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2}$
$v = \frac{\pi a\sqrt{3}}{T}$

(B) In $SHM$,the velocity $V$ of a particle 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 distances $x_1$ and $x_2$,we have:
$V_1^2 = \omega^2(a^2 - x_1^2) \dots (i)$
$V_2^2 = \omega^2(a^2 - x_2^2) \dots (ii)$
Subtracting equation $(ii)$ from $(i)$:
$V_1^2 - V_2^2 = \omega^2(a^2 - x_1^2 - a^2 + x_2^2)$
$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$
$\omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$
Since the time period $T = \frac{2\pi}{\omega}$,we have $\omega = \frac{2\pi}{T}$.
Substituting this into the equation for $\omega^2$:
$\left(\frac{2\pi}{T}\right)^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$
$T^2 = 4\pi^2 \left(\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}\right)$
$T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

(A) Given,amplitude $A = 3\,cm$ and displacement $x = 2\,cm$.
The velocity of a particle in simple harmonic motion is $v = \omega \sqrt{A^2 - x^2}$.
The magnitude of its acceleration is $a = \omega^2 x$.
Given that $|v| = |a|$,we have $\omega \sqrt{A^2 - x^2} = \omega^2 x$.
Dividing both sides by $\omega$ (assuming $\omega \neq 0$),we get $\sqrt{A^2 - x^2} = \omega x$.
Squaring both sides: $A^2 - x^2 = \omega^2 x^2$.
Substituting the values: $3^2 - 2^2 = \omega^2 (2^2) \Rightarrow 9 - 4 = 4\omega^2 \Rightarrow 5 = 4\omega^2$.
Thus,$\omega^2 = \frac{5}{4}$,which gives $\omega = \frac{\sqrt{5}}{2}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{5}/2} = \frac{4\pi}{\sqrt{5}}\,s$.

(B) The equation of motion is $x = A \sin(\omega t + \frac{\pi}{6})$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \frac{\pi}{6})$.
The maximum velocity $v_{max}$ is $A\omega$.
We are given that the velocity is half of its maximum velocity: $v = \frac{v_{max}}{2}$.
Substituting the expressions: $A\omega \cos(\omega t + \frac{\pi}{6}) = \frac{A\omega}{2}$.
This simplifies to $\cos(\omega t + \frac{\pi}{6}) = \frac{1}{2}$.
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $\omega t + \frac{\pi}{6} = \frac{\pi}{3}$.
Using $\omega = \frac{2\pi}{T}$,we get $\frac{2\pi}{T} t + \frac{\pi}{6} = \frac{\pi}{3}$.
Solving for $t$: $\frac{2\pi}{T} t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$t = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12}$.

(A) The displacement equation is $x = A \sin (2\pi t + \pi /3).$
Velocity $v$ is given by the derivative of displacement with respect to time: $v = \frac{dx}{dt} = A(2\pi) \cos (2\pi t + \pi /3).$
Velocity is maximum when the magnitude of the cosine term is maximum,which occurs at the mean position where $x = 0$.
Setting $x = 0$,we get $\sin (2\pi t + \pi /3) = 0$.
This implies the argument $(2\pi t + \pi /3) = n\pi$,where $n$ is an integer.
For $n = 0$,$2\pi t = -\pi /3$,which gives $t < 0$ (not possible).
For $n = 1$,$2\pi t + \pi /3 = \pi$.
$2\pi t = \pi - \pi /3 = 2\pi /3$.
$t = 1/3 \approx 0.33 \text{ s}$.
Thus,the first time after $t = 0$ when the velocity is maximum is $0.33 \text{ s}$.

(B) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a \omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Since $\omega = \frac{2 \pi}{T}$,we have $v_{\max} = a \times \frac{2 \pi}{T}$.
Rearranging for the period $T$,we get $T = \frac{2 \pi a}{v_{\max}}$.
Given $a = 7 \ mm = 7 \times 10^{-3} \ m$ and $v_{\max} = 4.4 \ m/s$.
Substituting the values: $T = \frac{2 \times 3.14 \times 7 \times 10^{-3}}{4.4}$.
$T = \frac{43.96 \times 10^{-3}}{4.4} \approx 9.99 \times 10^{-3} \ s \approx 0.01 \ s$.

(B) The displacement equation is given by $x = 2 \times 10^{-2} \cos(\pi t)$.
To find the velocity $v$, we differentiate the displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} [2 \times 10^{-2} \cos(\pi t)] = -2 \times 10^{-2} \pi \sin(\pi t)$.
The speed is the magnitude of velocity, $|v| = |2 \times 10^{-2} \pi \sin(\pi t)|$.
For the speed to be maximum, the value of $|\sin(\pi t)|$ must be $1$.
This occurs when $\sin(\pi t) = 1$ or $\sin(\pi t) = -1$.
The first time this occurs is when $\sin(\pi t) = 1$, which corresponds to $\pi t = \frac{\pi}{2}$.
Solving for $t$, we get $t = \frac{1}{2} = 0.5 \text{ s}$.

(B) Let the initial amplitude be $A$ and the new amplitude be $A'$. The velocity $v$ of a particle in simple harmonic motion at a distance $x$ from the equilibrium position is given by $v = \omega \sqrt{A^2 - x^2}$.
At $x = \frac{2A}{3}$,the initial velocity $v$ is $v = \omega \sqrt{A^2 - (\frac{2A}{3})^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\omega A \sqrt{5}}{3}$.
The new speed is $v' = 3v = \omega A \sqrt{5}$.
Using the formula for the new amplitude $A'$ at the same position $x = \frac{2A}{3}$:
$(v')^2 = \omega^2 (A'^2 - x^2)$
$(\omega A \sqrt{5})^2 = \omega^2 (A'^2 - (\frac{2A}{3})^2)$
$5A^2 = A'^2 - \frac{4A^2}{9}$
$A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$
$A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$.

(D) From the graph,the maximum distance from the ceiling is $100\ cm$ and the minimum distance is $30\ cm$.
These represent the two extreme positions of the oscillation.
The total distance between the extreme positions is $100\ cm - 30\ cm = 70\ cm$.
Since the total distance between extreme positions is $2A$,we have $2A = 70\ cm$,which gives the amplitude $A = 35\ cm$.
The mean position is at a distance of $30\ cm + 35\ cm = 65\ cm$ from the ceiling.
At $t = 0$,the mass is at the upper extreme position $(100\ cm)$.
At $t = \frac{1}{4}\ T$,the mass reaches the mean position $(65\ cm)$.
At $t = \frac{1}{2}\ T$,the mass reaches the lower extreme position $(30\ cm)$.
In simple harmonic motion,the speed is maximum at the mean position.
Therefore,the speed is maximum at $t = \frac{1}{4}\ T$.

(A) The displacement equation for $SHM$ is $x(t) = A \sin(\omega(t - t_0))$,where $t_0$ is the time at which the particle is at the equilibrium position.
Given $t_0 = 1 \, s$,so $x(t) = A \sin(\omega(t - 1))$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \, rad/s$.
The velocity is $v(t) = \frac{dx}{dt} = A\omega \cos(\omega(t - 1))$.
At $t = 2 \, s$,$v = 0.25 \, m/s = \frac{1}{4} \, m/s$.
Substituting the values: $\frac{1}{4} = A \left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}(2 - 1)\right)$.
$\frac{1}{4} = A \left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right)$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $\frac{1}{4} = A \left(\frac{\pi}{3}\right) \left(\frac{1}{2}\right)$.
$\frac{1}{4} = A \left(\frac{\pi}{6}\right)$.
$A = \frac{6}{4\pi} = \frac{3}{2\pi} \, m$.

(A) For a simple harmonic oscillator,the displacement is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity is given by $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
From the given graph,at $t = 0$,the velocity $v(0)$ is maximum (positive peak).
If the pendulum starts from the mean position $(x=0)$,the velocity is maximum at $t=0$,which matches the graph.
If the pendulum starts from the extreme position $(x=A)$,the velocity is $0$ at $t=0$,which does not match the graph.
Therefore,the graph represents the motion of a simple pendulum starting from the mean position.