Velocity of Simple Harmonic Motion Questions in English

(B) In simple harmonic motion $(SHM)$,the maximum velocity of an object is given by:
$v_{\max} = A\omega$
From this,we can express the amplitude $A$ as:
$A = \frac{v_{\max}}{\omega} \quad ...(i)$
where $A$ is the amplitude and $\omega$ is the angular frequency.
The maximum acceleration of an object performing $SHM$ is given by:
$a_{\max} = \omega^2 A$
Substituting the value of $A$ from equation $(i)$ into the expression for $a_{\max}$:
$a_{\max} = \omega^2 \left( \frac{v_{\max}}{\omega} \right)$
$a_{\max} = \omega v_{\max}$

(B) The equation of a particle executing simple harmonic motion is given by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$.
The velocity $v$ of the particle is the derivative of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt} \left[ A \cos \left(\omega t + \frac{\pi}{4}\right) \right] = -A \omega \sin \left(\omega t + \frac{\pi}{4}\right)$.
The speed $|v|$ is maximum when the magnitude of the sine function is $1$,i.e.,$\sin \left(\omega t + \frac{\pi}{4}\right) = -1$ or $1$.
For the first positive time $t$,we consider $\sin \left(\omega t + \frac{\pi}{4}\right) = -1 = \sin \left(\frac{3\pi}{2}\right)$.
Thus,$\omega t + \frac{\pi}{4} = \frac{3\pi}{2} \Rightarrow \omega t = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{5\pi}{4} \Rightarrow t = \frac{5\pi}{4\omega}$.
However,checking the options provided,if we consider the condition for maximum speed magnitude $|v| = A\omega$,we require $\sin^2(\omega t + \pi/4) = 1$.
If we set $\omega t + \pi/4 = \pi/2$,we get $t = \pi/(4\omega)$,which corresponds to the particle passing through the equilibrium position with maximum speed in the negative direction.

(B) In simple harmonic motion $(SHM)$,the velocity of a particle is zero at the extreme positions.
Let the displacement be $x = A \sin(\omega t)$. The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.
The velocity is zero when $\cos(\omega t) = 0$,which occurs at $t = \frac{T}{4}, \frac{3T}{4}, \dots$
The time interval between two consecutive points of zero velocity (i.e.,between two extreme positions) is the time taken to travel from one extreme to the other,which is half the time period,$\frac{T}{2}$.
Given,$\frac{T}{2} = 0.5 \text{ s}$.
Therefore,$T = 1 \text{ s}$.
The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T}$.
Substituting $T = 1 \text{ s}$,we get $\omega = \frac{2\pi}{1} = 2\pi \text{ rad s}^{-1}$.

(A) Given,maximum speed of $SHM$ is $v_{\max} = A\omega = 40 \,ms^{-1}$ ... $(i)$
Maximum acceleration of $SHM$ is $a_{\max} = A\omega^2 = 60 \,ms^{-2}$ ... (ii)
Dividing Eq. (ii) by Eq. $(i)$,we get:
$\frac{A\omega^2}{A\omega} = \frac{60}{40}$
$\omega = 1.5 \,rad/s = \frac{3}{2} \,rad/s$
We know that the time period $T$ is related to angular frequency $\omega$ by the formula:
$T = \frac{2\pi}{\omega}$
Substituting the value of $\omega$:
$T = \frac{2\pi}{3/2} = \frac{4\pi}{3} \,s$
Hence,the time period of the oscillation is $\frac{4\pi}{3} \,s$.

(B) In simple harmonic motion $(SHM)$, the maximum velocity is given by $v_{\text{max}} = \omega a$, where $\omega$ is the angular frequency and $a$ is the amplitude.
Given, $v_{\text{max}} = 6.28 \text{ cm s}^{-1}$.
The length of the path is the total distance covered by the particle in one oscillation, which is equal to $2a$.
So, $2a = 8 \text{ cm}$, which implies $a = 4 \text{ cm}$.
Using the formula $v_{\text{max}} = \omega a$, we have $\omega = \frac{v_{\text{max}}}{a} = \frac{6.28}{4} \text{ rad s}^{-1}$.
Since $\omega = \frac{2\pi}{T}$, we can write $\frac{2\pi}{T} = \frac{6.28}{4}$.
Substituting $\pi \approx 3.14$, we get $\frac{2 \times 3.14}{T} = \frac{6.28}{4}$.
$\frac{6.28}{T} = \frac{6.28}{4}$.
Therefore, $T = 4 \text{ s}$.

(B) The points $A$ and $B$ are the extreme positions of the $SHM$ because the velocity is zero at these points.
The distance between the extreme positions $A$ and $B$ is $L = b - a$.
The amplitude $A_{amp}$ of the $SHM$ is half the distance between the extreme positions:
$A_{amp} = \frac{b - a}{2}$
The equilibrium position (mean position) is at the midpoint of $A$ and $B$,which is at a distance of $\frac{a + b}{2}$ from $O$.
The velocity of a particle in $SHM$ at a displacement $x$ from the mean position is given by $v = \omega \sqrt{A_{amp}^2 - x^2}$.
At the midpoint between $A$ and $B$,the particle is at the mean position,so the displacement $x = 0$.
Thus,the velocity at the midpoint is the maximum velocity $v_{max} = \omega A_{amp}$.
Given $v_{max} = v$,we have $v = \omega \left(\frac{b - a}{2}\right)$.
Solving for angular frequency $\omega$:
$\omega = \frac{2v}{b - a}$
The time period $T$ is given by:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{\left(\frac{2v}{b - a}\right)} = \frac{\pi(b - a)}{v}$

(B) In simple harmonic motion,the force is given by $F = -kx = -m\omega^2 x$. The maximum force is $F_{max} = m\omega^2 A$.
Given that the force at a position $x$ is $86.6 \%$ of $F_{max}$,we have $F = 0.866 F_{max} = \frac{\sqrt{3}}{2} F_{max}$.
Since $F = m\omega^2 x$ and $F_{max} = m\omega^2 A$,we get $x = \frac{\sqrt{3}}{2} A$.
The velocity of a particle in $SHM$ at position $x$ is $v = \omega \sqrt{A^2 - x^2}$.
Substituting $x = \frac{\sqrt{3}}{2} A$,we get $v = \omega \sqrt{A^2 - (\frac{\sqrt{3}}{2} A)^2} = \omega \sqrt{A^2 - \frac{3}{4} A^2} = \omega \sqrt{\frac{1}{4} A^2} = \frac{1}{2} \omega A$.
The maximum velocity is $v_{max} = \omega A$.
Therefore,the ratio of velocity at that point to the maximum velocity is $\frac{v}{v_{max}} = \frac{\frac{1}{2} \omega A}{\omega A} = \frac{1}{2}$.

(D) The equation of motion for a particle in simple harmonic motion is $F = -ky$. Given $F + 0.04 \pi^2 y = 0$, we have $F = -0.04 \pi^2 y$. Comparing this with $F = -ky$, the force constant $k = 0.04 \pi^2 \text{ N/m}$.
The mass of the particle is $m = 90 \text{ g} = 0.09 \text{ kg}$.
The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{0.04 \pi^2}{0.09}} = \sqrt{\frac{4 \pi^2}{9}} = \frac{2 \pi}{3} \text{ rad/s}$.
The amplitude $A = \frac{6}{\pi} \text{ m}$.
The maximum velocity $v_{\text{max}}$ is given by $v_{\text{max}} = A \omega$.
Substituting the values, $v_{\text{max}} = \left( \frac{6}{\pi} \right) \times \left( \frac{2 \pi}{3} \right) = 4 \text{ m/s}$.

(B) The equation of motion for a particle in $SHM$ passing through the origin at $t=2 \ s$ is $x(t) = A \sin(\omega(t - 2))$.
Given $T = 16 \ s$,the angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{16} = \frac{\pi}{8} \ rad/s$.
The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos(\omega(t - 2))$.
At $t = 4 \ s$,$v = 4 \ m/s$:
$4 = A \left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}(4 - 2)\right)$
$4 = A \left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{4}\right)$
$4 = A \left(\frac{\pi}{8}\right) \left(\frac{1}{\sqrt{2}}\right)$
$A = \frac{4 \times 8 \times \sqrt{2}}{\pi} = \frac{32 \sqrt{2}}{\pi} \ m$.

(B) The acceleration of a particle in $\text{S.H.M.}$ is given by $a = -\omega^2 x$. Taking the magnitude,$|a| = \omega^2 |x|$.
Given $x = 1 \ cm$ and $a = 3 \ cm s^{-2}$,we have $3 = \omega^2 (1)$,which implies $\omega^2 = 3 \ s^{-2}$.
The velocity of a particle in $\text{S.H.M.}$ is given by $v = \omega \sqrt{A^2 - x^2}$.
Given $v = 6 \ cm s^{-1}$ at $x = 2 \ cm$,we substitute the values:
$6 = \sqrt{3} \sqrt{A^2 - 2^2}$.
Squaring both sides:
$36 = 3 (A^2 - 4)$.
Dividing by $3$:
$12 = A^2 - 4$.
$A^2 = 16$.
$A = 4 \ cm$.

(D) Given the equation relating velocity $v$ and displacement $x$ for a particle in simple harmonic motion $(SHM)$:
$4v^2 = 25 - x^2$
Dividing by $4$,we get:
$v^2 = \frac{1}{4}(25 - x^2) = \frac{1}{4}(5^2 - x^2)$
Taking the square root on both sides:
$v = \frac{1}{2}\sqrt{5^2 - x^2} \quad ... (i)$
We know that the standard equation for velocity in $SHM$ is:
$v = \omega\sqrt{A^2 - x^2} \quad ... (ii)$
Comparing equations $(i)$ and $(ii)$,we find:
Angular frequency $\omega = \frac{1}{2} \ rad/s$
Amplitude $A = 5 \ m$
The time period $T$ is given by the formula:
$T = \frac{2\pi}{\omega}$
Substituting the value of $\omega$:
$T = \frac{2\pi}{1/2} = 4\pi \ s$

(B) Let the amplitude be $A$. The angular frequencies are $\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{3}$ and $\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{6} = \frac{\pi}{3}$.
Since both start from the origin at $t=0$,their displacements are $x_1 = A \sin(\omega_1 t)$ and $x_2 = A \sin(\omega_2 t)$.
When they meet,$x_1 = x_2$,so $\sin(\omega_1 t) = \sin(\omega_2 t)$.
For the first meeting after $t=0$,$\omega_1 t = \pi - \omega_2 t$,which gives $t = \frac{\pi}{\omega_1 + \omega_2} = \frac{\pi}{\frac{2\pi}{3} + \frac{\pi}{3}} = 1 \ s$.
The velocity of a particle in $SHM$ is $v = A\omega \cos(\omega t)$.
The ratio of velocities is $\frac{v_P}{v_Q} = \frac{A\omega_1 \cos(\omega_1 t)}{A\omega_2 \cos(\omega_2 t)} = \frac{(2\pi/3) \cos(2\pi/3 \cdot 1)}{(\pi/3) \cos(\pi/3 \cdot 1)} = \frac{2 \cos(120^\circ)}{\cos(60^\circ)} = \frac{2(-1/2)}{1/2} = -2$.
The magnitude ratio is $2:1$.

(D) Given, amplitude, $A = 0.2 \,cm = 2 \times 10^{-3} \,m$.
Velocity at the mean position in simple harmonic motion is the maximum velocity, $v_{\text{max}} = 5 \,m/s$.
We know that the maximum velocity is given by the formula, $v_{\text{max}} = A \omega$.
Therefore, the angular frequency $\omega$ is given by, $\omega = \frac{v_{\text{max}}}{A}$.
Substituting the values, $\omega = \frac{5}{2 \times 10^{-3}} = 2.5 \times 10^3 = 2500 \,rad/s$.

(C) The equation for the position of a particle in simple harmonic motion starting from the mean position is $y(t) = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.
The distance travelled in the first second ($t=0$ to $t=1$) is $y_1 = A \sin(\frac{\pi}{4} \times 1) = A \times \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}}$.
The position at $t=2$ seconds is $y(2) = A \sin(\frac{\pi}{4} \times 2) = A \sin(\frac{\pi}{2}) = A$.
The distance travelled in the second second ($t=1$ to $t=2$) is $y_2 = y(2) - y(1) = A - \frac{A}{\sqrt{2}} = A(1 - \frac{1}{\sqrt{2}})$.
The ratio of the distances is $\frac{y_1}{y_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1/\sqrt{2}}{(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.

(C) For a simple harmonic motion,the displacement is given by $y = A \sin(\omega t)$ and velocity is $v = \frac{dy}{dt} = A \omega \cos(\omega t)$.
Rearranging these,we get $\frac{y}{A} = \sin(\omega t)$ and $\frac{v}{A \omega} = \cos(\omega t)$.
Squaring and adding,we get $\frac{v^2}{(A \omega)^2} + \frac{y^2}{A^2} = 1$,which represents an ellipse.
The major axis along the $X$-axis (velocity) is $2A \omega$ and the minor axis along the $Y$-axis (displacement) is $2A$.
The ratio of the major axis to the minor axis is $\frac{2A \omega}{2A} = \omega$.
Given that the ratio is $20 \pi$,we have $\omega = 20 \pi$.
Since $\omega = 2 \pi f$,we have $2 \pi f = 20 \pi$.
Therefore,$f = 10 \ Hz$.

(B) The speed of a particle executing simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude,$\omega$ is the angular frequency,and $x$ is the displacement from the equilibrium position.
Squaring both sides,we get $v^2 = \omega^2 (a^2 - x^2)$.
For the two given cases:
$v_1^2 = \omega^2 (a^2 - x_1^2) \implies 13^2 = \omega^2 (a^2 - 3^2) \implies 169 = \omega^2 (a^2 - 9) \quad (1)$
$v_2^2 = \omega^2 (a^2 - x_2^2) \implies 12^2 = \omega^2 (a^2 - 5^2) \implies 144 = \omega^2 (a^2 - 25) \quad (2)$
Subtracting equation $(2)$ from $(1)$:
$169 - 144 = \omega^2 (a^2 - 9 - a^2 + 25)$
$25 = \omega^2 (16)$
$\omega^2 = \frac{25}{16} \implies \omega = \frac{5}{4} \ rad/s$.
The frequency $f$ is related to angular frequency by $\omega = 2 \pi f$.
Therefore,$f = \frac{\omega}{2 \pi} = \frac{5/4}{2 \pi} = \frac{5}{8 \pi} \ Hz$.

(A) The angular frequency of the kinetic energy oscillation is given as $\omega_k = 176 \ rad/s$.
For a simple harmonic oscillator,the kinetic energy $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t + \phi)$.
Using the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$,we get $K = \frac{1}{4} m \omega^2 A^2 (1 - \cos(2\omega t + 2\phi))$.
This shows that the kinetic energy oscillates with an angular frequency $\omega_k = 2\omega$,where $\omega$ is the angular frequency of the oscillator.
Given $\omega_k = 176 \ rad/s$,we have $2\omega = 176 \ rad/s$,which implies $\omega = 88 \ rad/s$.
The frequency of the oscillator is $f = \frac{\omega}{2\pi} = \frac{88}{2 \times (22/7)} = \frac{88 \times 7}{44} = 2 \times 7 = 14 \ Hz$.

(A) The standard equation for velocity in simple harmonic motion $(SHM)$ is $v^2 = \omega^2 (A^2 - x^2)$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Given the equation $v^2 = 50 - x^2$,we can rewrite it as $v^2 = 1(50 - x^2)$.
Comparing this with the standard equation,we get $\omega^2 = 1$,which implies $\omega = 1 \ \text{rad/s}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi \ \text{s}$.
Using the approximation $\pi \approx \frac{22}{7}$,we get $T = 2 \times \frac{22}{7} = \frac{44}{7} \ \text{s}$.
According to the problem,$T = \frac{x}{7} \ \text{s}$.
Equating the two expressions for $T$: $\frac{x}{7} = \frac{44}{7}$.
Therefore,$x = 44$.