364258
Two simple harmonic motions of angular frequency \({\rm{100}}\) and \(1000\,rad\,{s^{ - 1}}\) have the same displacement amplitude. The ratio of their maximum acceleration is
1 \(1: 10\)
2 \(1: 10^{2}\)
3 \(1: 10^{3}\)
4 \(1: 10^{4}\)
Explanation:
Acceleration of simple harmonic motion is \(a_{\max }=-\omega^{2} A\) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{\omega_{1}^{2}}{\omega_{2}^{2}} \quad\) (as \(A\) remains the same) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{(100)^{2}}{(1000)^{2}}=\left(\dfrac{1}{10}\right)^{2}=1: 10^{2}\)
PHXI14:OSCILLATIONS
364259
A particle performs simple harmonic motion with amplitude \(A\). Its speed is tripled at the instant when it is at a distance \(\dfrac{2 A}{3}\) from equilibrium position. The new amplitude of the motion is :
1 \(3\;A\)
2 \(A\sqrt 3 \)
3 \(\dfrac{7 A}{3}\)
4 \(\dfrac{A}{3} \sqrt{41}\)
Explanation:
We know that \(v=\omega \sqrt{A^{2}-x^{2}}\) Initially \(v=\omega \sqrt{A^{2}-\left(\dfrac{2 A}{3}\right)^{2}}\) After increasing the velocity \(3v = \omega \sqrt {{{A'}^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} \) Where \(A^{\prime}\) is the final amplitude. \(\frac{1}{3} = \sqrt {\frac{{{A^\prime } - {{\left( {\frac{{2A}}{3}} \right)}^2}}}{{{A^{\prime 2}} - {{\left( {\frac{{2A}}{3}} \right)}^2}}}} \) \(9\left[ {{A^2} - \frac{{4{A^2}}}{9}} \right] = {A^{\prime 2}} - \frac{{4{A^2}}}{9}\) \( \Rightarrow {A^\prime } = \frac{{7A}}{3}\)
PHXI14:OSCILLATIONS
364260
The equation of a simple harmonic wave is give by \(y=3 \sin \dfrac{\pi}{2}(50 t-x)\), where \(x\) and \(y\) are in meters and \(t\) is in seconds. The ratio of maximum particle velocity to the wave velocity is
1 \(2 \pi\)
2 \(\dfrac{3}{2} \pi\)
3 \(3 \pi\)
4 \(\dfrac{2}{3} \pi\)
Explanation:
We now that \(v_{\max }=a \omega\) and \(v=\lambda f\) \(\begin{aligned}& \therefore \dfrac{v_{\max }}{v}=\dfrac{a \omega}{f \lambda}=\dfrac{a(2 \pi f)}{f \lambda}=\dfrac{2 \pi a}{\lambda} \\& \dfrac{2 \pi a}{2 \pi / k}=k a=\dfrac{\pi}{2} \times 3=\dfrac{3 \pi}{2}\end{aligned}\)
PHXI14:OSCILLATIONS
364261
The amplitude of a particle executing \(S H M\) is \(4\;cm.\) At the mean position, the speed of the particle is \(16\;cm/s.\) The distance of the particle from the mean position at which the speed of the particle becomes \(8\sqrt 3 \;cm/s\) will be
364258
Two simple harmonic motions of angular frequency \({\rm{100}}\) and \(1000\,rad\,{s^{ - 1}}\) have the same displacement amplitude. The ratio of their maximum acceleration is
1 \(1: 10\)
2 \(1: 10^{2}\)
3 \(1: 10^{3}\)
4 \(1: 10^{4}\)
Explanation:
Acceleration of simple harmonic motion is \(a_{\max }=-\omega^{2} A\) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{\omega_{1}^{2}}{\omega_{2}^{2}} \quad\) (as \(A\) remains the same) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{(100)^{2}}{(1000)^{2}}=\left(\dfrac{1}{10}\right)^{2}=1: 10^{2}\)
PHXI14:OSCILLATIONS
364259
A particle performs simple harmonic motion with amplitude \(A\). Its speed is tripled at the instant when it is at a distance \(\dfrac{2 A}{3}\) from equilibrium position. The new amplitude of the motion is :
1 \(3\;A\)
2 \(A\sqrt 3 \)
3 \(\dfrac{7 A}{3}\)
4 \(\dfrac{A}{3} \sqrt{41}\)
Explanation:
We know that \(v=\omega \sqrt{A^{2}-x^{2}}\) Initially \(v=\omega \sqrt{A^{2}-\left(\dfrac{2 A}{3}\right)^{2}}\) After increasing the velocity \(3v = \omega \sqrt {{{A'}^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} \) Where \(A^{\prime}\) is the final amplitude. \(\frac{1}{3} = \sqrt {\frac{{{A^\prime } - {{\left( {\frac{{2A}}{3}} \right)}^2}}}{{{A^{\prime 2}} - {{\left( {\frac{{2A}}{3}} \right)}^2}}}} \) \(9\left[ {{A^2} - \frac{{4{A^2}}}{9}} \right] = {A^{\prime 2}} - \frac{{4{A^2}}}{9}\) \( \Rightarrow {A^\prime } = \frac{{7A}}{3}\)
PHXI14:OSCILLATIONS
364260
The equation of a simple harmonic wave is give by \(y=3 \sin \dfrac{\pi}{2}(50 t-x)\), where \(x\) and \(y\) are in meters and \(t\) is in seconds. The ratio of maximum particle velocity to the wave velocity is
1 \(2 \pi\)
2 \(\dfrac{3}{2} \pi\)
3 \(3 \pi\)
4 \(\dfrac{2}{3} \pi\)
Explanation:
We now that \(v_{\max }=a \omega\) and \(v=\lambda f\) \(\begin{aligned}& \therefore \dfrac{v_{\max }}{v}=\dfrac{a \omega}{f \lambda}=\dfrac{a(2 \pi f)}{f \lambda}=\dfrac{2 \pi a}{\lambda} \\& \dfrac{2 \pi a}{2 \pi / k}=k a=\dfrac{\pi}{2} \times 3=\dfrac{3 \pi}{2}\end{aligned}\)
PHXI14:OSCILLATIONS
364261
The amplitude of a particle executing \(S H M\) is \(4\;cm.\) At the mean position, the speed of the particle is \(16\;cm/s.\) The distance of the particle from the mean position at which the speed of the particle becomes \(8\sqrt 3 \;cm/s\) will be
364258
Two simple harmonic motions of angular frequency \({\rm{100}}\) and \(1000\,rad\,{s^{ - 1}}\) have the same displacement amplitude. The ratio of their maximum acceleration is
1 \(1: 10\)
2 \(1: 10^{2}\)
3 \(1: 10^{3}\)
4 \(1: 10^{4}\)
Explanation:
Acceleration of simple harmonic motion is \(a_{\max }=-\omega^{2} A\) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{\omega_{1}^{2}}{\omega_{2}^{2}} \quad\) (as \(A\) remains the same) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{(100)^{2}}{(1000)^{2}}=\left(\dfrac{1}{10}\right)^{2}=1: 10^{2}\)
PHXI14:OSCILLATIONS
364259
A particle performs simple harmonic motion with amplitude \(A\). Its speed is tripled at the instant when it is at a distance \(\dfrac{2 A}{3}\) from equilibrium position. The new amplitude of the motion is :
1 \(3\;A\)
2 \(A\sqrt 3 \)
3 \(\dfrac{7 A}{3}\)
4 \(\dfrac{A}{3} \sqrt{41}\)
Explanation:
We know that \(v=\omega \sqrt{A^{2}-x^{2}}\) Initially \(v=\omega \sqrt{A^{2}-\left(\dfrac{2 A}{3}\right)^{2}}\) After increasing the velocity \(3v = \omega \sqrt {{{A'}^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} \) Where \(A^{\prime}\) is the final amplitude. \(\frac{1}{3} = \sqrt {\frac{{{A^\prime } - {{\left( {\frac{{2A}}{3}} \right)}^2}}}{{{A^{\prime 2}} - {{\left( {\frac{{2A}}{3}} \right)}^2}}}} \) \(9\left[ {{A^2} - \frac{{4{A^2}}}{9}} \right] = {A^{\prime 2}} - \frac{{4{A^2}}}{9}\) \( \Rightarrow {A^\prime } = \frac{{7A}}{3}\)
PHXI14:OSCILLATIONS
364260
The equation of a simple harmonic wave is give by \(y=3 \sin \dfrac{\pi}{2}(50 t-x)\), where \(x\) and \(y\) are in meters and \(t\) is in seconds. The ratio of maximum particle velocity to the wave velocity is
1 \(2 \pi\)
2 \(\dfrac{3}{2} \pi\)
3 \(3 \pi\)
4 \(\dfrac{2}{3} \pi\)
Explanation:
We now that \(v_{\max }=a \omega\) and \(v=\lambda f\) \(\begin{aligned}& \therefore \dfrac{v_{\max }}{v}=\dfrac{a \omega}{f \lambda}=\dfrac{a(2 \pi f)}{f \lambda}=\dfrac{2 \pi a}{\lambda} \\& \dfrac{2 \pi a}{2 \pi / k}=k a=\dfrac{\pi}{2} \times 3=\dfrac{3 \pi}{2}\end{aligned}\)
PHXI14:OSCILLATIONS
364261
The amplitude of a particle executing \(S H M\) is \(4\;cm.\) At the mean position, the speed of the particle is \(16\;cm/s.\) The distance of the particle from the mean position at which the speed of the particle becomes \(8\sqrt 3 \;cm/s\) will be
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PHXI14:OSCILLATIONS
364258
Two simple harmonic motions of angular frequency \({\rm{100}}\) and \(1000\,rad\,{s^{ - 1}}\) have the same displacement amplitude. The ratio of their maximum acceleration is
1 \(1: 10\)
2 \(1: 10^{2}\)
3 \(1: 10^{3}\)
4 \(1: 10^{4}\)
Explanation:
Acceleration of simple harmonic motion is \(a_{\max }=-\omega^{2} A\) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{\omega_{1}^{2}}{\omega_{2}^{2}} \quad\) (as \(A\) remains the same) or \(\dfrac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\dfrac{(100)^{2}}{(1000)^{2}}=\left(\dfrac{1}{10}\right)^{2}=1: 10^{2}\)
PHXI14:OSCILLATIONS
364259
A particle performs simple harmonic motion with amplitude \(A\). Its speed is tripled at the instant when it is at a distance \(\dfrac{2 A}{3}\) from equilibrium position. The new amplitude of the motion is :
1 \(3\;A\)
2 \(A\sqrt 3 \)
3 \(\dfrac{7 A}{3}\)
4 \(\dfrac{A}{3} \sqrt{41}\)
Explanation:
We know that \(v=\omega \sqrt{A^{2}-x^{2}}\) Initially \(v=\omega \sqrt{A^{2}-\left(\dfrac{2 A}{3}\right)^{2}}\) After increasing the velocity \(3v = \omega \sqrt {{{A'}^2} - {{\left( {\frac{{2A}}{3}} \right)}^2}} \) Where \(A^{\prime}\) is the final amplitude. \(\frac{1}{3} = \sqrt {\frac{{{A^\prime } - {{\left( {\frac{{2A}}{3}} \right)}^2}}}{{{A^{\prime 2}} - {{\left( {\frac{{2A}}{3}} \right)}^2}}}} \) \(9\left[ {{A^2} - \frac{{4{A^2}}}{9}} \right] = {A^{\prime 2}} - \frac{{4{A^2}}}{9}\) \( \Rightarrow {A^\prime } = \frac{{7A}}{3}\)
PHXI14:OSCILLATIONS
364260
The equation of a simple harmonic wave is give by \(y=3 \sin \dfrac{\pi}{2}(50 t-x)\), where \(x\) and \(y\) are in meters and \(t\) is in seconds. The ratio of maximum particle velocity to the wave velocity is
1 \(2 \pi\)
2 \(\dfrac{3}{2} \pi\)
3 \(3 \pi\)
4 \(\dfrac{2}{3} \pi\)
Explanation:
We now that \(v_{\max }=a \omega\) and \(v=\lambda f\) \(\begin{aligned}& \therefore \dfrac{v_{\max }}{v}=\dfrac{a \omega}{f \lambda}=\dfrac{a(2 \pi f)}{f \lambda}=\dfrac{2 \pi a}{\lambda} \\& \dfrac{2 \pi a}{2 \pi / k}=k a=\dfrac{\pi}{2} \times 3=\dfrac{3 \pi}{2}\end{aligned}\)
PHXI14:OSCILLATIONS
364261
The amplitude of a particle executing \(S H M\) is \(4\;cm.\) At the mean position, the speed of the particle is \(16\;cm/s.\) The distance of the particle from the mean position at which the speed of the particle becomes \(8\sqrt 3 \;cm/s\) will be
