140371
The displacement of a particle of mass $2 \mathrm{~g}$ executing $S H M$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here, $y$ is in metres and $t$ is in seconds. The kinetic energy of the particle, when $t=\frac{T}{4}$ is
1 $0.4 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $0.3 \mathrm{~J}$
Explanation:
D Displacement equation in S.H.M - $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$ $y=a \sin (\omega t+\phi)$ On comparing equation, we get - $\omega=4, \quad \mathrm{a}=5$ $\mathrm{~T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2} \text { second }$ $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\pi}{2 \times 4}=\frac{\pi}{8} \text { second }$ $\mathrm{y}=5 \sin \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ On differentiating both side wrt time $\frac{\mathrm{dy}}{\mathrm{dt}}=20 \cos \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ $\operatorname{Velocity}(v)=20 \cos \left(4 t+\frac{\pi}{3}\right)$ $\mathrm{v}=20 \cos \left(4 \times \frac{\pi}{8}+\frac{\pi}{3}\right)=20 \cos 150^{\circ}\left(\because \mathrm{t}=\frac{\pi}{8}\right)$ $\mathrm{v}=-20 \cos 30^{\circ}=\frac{-20 \sqrt{3}}{2} \mathrm{~m} / \mathrm{sec}$ Kinetic energy $($ K.E. $)=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \times \frac{2}{1000}(-10 \sqrt{3})^{2}$ $\text { K.E. }=\frac{1}{2} \times \frac{2}{1000} \times 100 \times 3=\frac{1 \times 300}{1000}$ $\text { K.E. }=0.3 \mathrm{~J}$
AP EAMCET (23.04.2018) Shift-2
Oscillations
140372
The potential energy of a simple harmonic oscillator of mass $2 \mathrm{~kg}$ at its mean position is 5 $\mathrm{J}$. If its total energy is $9 \mathrm{~J}$ and amplitude is 1 $\mathrm{cm}$, then its time period is
1 $\frac{\pi}{100} \mathrm{~s}$
2 $\frac{\pi}{50} \mathrm{~s}$
3 $\frac{\pi}{20} \mathrm{~s}$
4 $\frac{\pi}{10} \mathrm{~s}$
Explanation:
A Given, $\mathrm{m}=2 \mathrm{~kg}, \mathrm{a}=0.01 \mathrm{~m}$ Potential energy at mean position $=5 \mathrm{~J}$ Total energy $=9 \mathrm{~J}$ Total energy $=(\text { Potential energy })_{\text {mean }}+(\text { K.E })_{\max }$ $9=5+(\mathrm{K} . \mathrm{E})_{\max }$ $(\mathrm{K} . \mathrm{E})_{\max }=4 \mathrm{~J}$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{~J}$ $\mathrm{v}_{\text {max }}=\sqrt{\frac{4 \times 2}{\mathrm{~m}}}=\sqrt{\frac{4 \times 2}{2}}=2 \mathrm{~m} / \mathrm{sec}$ Maximum velocity $=\mathrm{a} \omega$ $2=0.01 \omega$ $\omega=200 \mathrm{rad} / \mathrm{sec}$ We know that $\mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \times \pi}{200}=\frac{\pi}{100} \mathrm{~s}$
AP EAMCET (22.04.2018) Shift-1
Oscillations
140373
A particle move according to the law $x=\operatorname{rcos} \frac{\pi t}{2}$. The distance covered by it in the time interval between $t=0$ and $t=3 \mathrm{~s}$ is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $3 \mathrm{r}$
4 $4 \mathrm{r}$
Explanation:
C Given that, $x=\operatorname{rcos} \frac{\pi t}{2}$ Comparing the given equation, we get - $\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}$ $\text { Then, } \omega=\frac{\pi}{2}$ $\text { or } \frac{2 \pi}{\mathrm{T}}=\frac{\pi}{2}$ $\mathrm{~T}=4 \mathrm{~s}$ The given time $(\mathrm{t})=3 \mathrm{~s}$ is $\frac{3 \mathrm{~T}}{4}$ At time $(\mathrm{t})=0$ particle is at $\mathrm{x}=\mathrm{A}$ (at extreme position) and at $\mathrm{t}=3 \mathrm{~s}=\frac{3 \mathrm{~T}}{4}$ it will be at mean position, $\mathrm{x}=0$. $\therefore$ Distance covered will be 3 r.
JCECE-2017
Oscillations
140374
A particle is executing simple harmonic motion. Its displacement to amplitude ratio when its kinetic energy is $84 \%$ of total energy is
1 $1: 16$
2 $2: 5$
3 $4: 25$
4 $21: 25$
Explanation:
B Let the total energy be, $\mathrm{E}_{\text {total }}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ And kinetic energy (K.E.) $=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ Now, it displacement to amplitude ratio is given as $\frac{K E}{E_{\text {total }}}=84 \%$ $\frac{K E}{E_{\text {total }}}=\frac{84}{100}$ $\frac{\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)}{\frac{1}{2} m \omega^{2} A^{2}}=\frac{84}{100}$ $\frac{\left(A^{2}-x^{2}\right)}{A^{2}}=\frac{84}{100}$ $\frac{1-\frac{x^{2}}{A^{2}}=\frac{84}{100}}{\frac{x^{2}}{A^{2}}=1-\frac{84}{100}=\frac{100-84}{100}}$ $\frac{x^{2}}{A^{2}}=\frac{16}{100}$ $\frac{x}{A}=\sqrt{\frac{16}{100}}$ $\frac{x}{A}=\frac{4}{10}=\frac{2}{5}$ $\frac{x}{A}=2: 5$ Hence, $\mathrm{x}: \mathrm{A}=2: 5$
140371
The displacement of a particle of mass $2 \mathrm{~g}$ executing $S H M$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here, $y$ is in metres and $t$ is in seconds. The kinetic energy of the particle, when $t=\frac{T}{4}$ is
1 $0.4 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $0.3 \mathrm{~J}$
Explanation:
D Displacement equation in S.H.M - $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$ $y=a \sin (\omega t+\phi)$ On comparing equation, we get - $\omega=4, \quad \mathrm{a}=5$ $\mathrm{~T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2} \text { second }$ $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\pi}{2 \times 4}=\frac{\pi}{8} \text { second }$ $\mathrm{y}=5 \sin \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ On differentiating both side wrt time $\frac{\mathrm{dy}}{\mathrm{dt}}=20 \cos \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ $\operatorname{Velocity}(v)=20 \cos \left(4 t+\frac{\pi}{3}\right)$ $\mathrm{v}=20 \cos \left(4 \times \frac{\pi}{8}+\frac{\pi}{3}\right)=20 \cos 150^{\circ}\left(\because \mathrm{t}=\frac{\pi}{8}\right)$ $\mathrm{v}=-20 \cos 30^{\circ}=\frac{-20 \sqrt{3}}{2} \mathrm{~m} / \mathrm{sec}$ Kinetic energy $($ K.E. $)=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \times \frac{2}{1000}(-10 \sqrt{3})^{2}$ $\text { K.E. }=\frac{1}{2} \times \frac{2}{1000} \times 100 \times 3=\frac{1 \times 300}{1000}$ $\text { K.E. }=0.3 \mathrm{~J}$
AP EAMCET (23.04.2018) Shift-2
Oscillations
140372
The potential energy of a simple harmonic oscillator of mass $2 \mathrm{~kg}$ at its mean position is 5 $\mathrm{J}$. If its total energy is $9 \mathrm{~J}$ and amplitude is 1 $\mathrm{cm}$, then its time period is
1 $\frac{\pi}{100} \mathrm{~s}$
2 $\frac{\pi}{50} \mathrm{~s}$
3 $\frac{\pi}{20} \mathrm{~s}$
4 $\frac{\pi}{10} \mathrm{~s}$
Explanation:
A Given, $\mathrm{m}=2 \mathrm{~kg}, \mathrm{a}=0.01 \mathrm{~m}$ Potential energy at mean position $=5 \mathrm{~J}$ Total energy $=9 \mathrm{~J}$ Total energy $=(\text { Potential energy })_{\text {mean }}+(\text { K.E })_{\max }$ $9=5+(\mathrm{K} . \mathrm{E})_{\max }$ $(\mathrm{K} . \mathrm{E})_{\max }=4 \mathrm{~J}$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{~J}$ $\mathrm{v}_{\text {max }}=\sqrt{\frac{4 \times 2}{\mathrm{~m}}}=\sqrt{\frac{4 \times 2}{2}}=2 \mathrm{~m} / \mathrm{sec}$ Maximum velocity $=\mathrm{a} \omega$ $2=0.01 \omega$ $\omega=200 \mathrm{rad} / \mathrm{sec}$ We know that $\mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \times \pi}{200}=\frac{\pi}{100} \mathrm{~s}$
AP EAMCET (22.04.2018) Shift-1
Oscillations
140373
A particle move according to the law $x=\operatorname{rcos} \frac{\pi t}{2}$. The distance covered by it in the time interval between $t=0$ and $t=3 \mathrm{~s}$ is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $3 \mathrm{r}$
4 $4 \mathrm{r}$
Explanation:
C Given that, $x=\operatorname{rcos} \frac{\pi t}{2}$ Comparing the given equation, we get - $\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}$ $\text { Then, } \omega=\frac{\pi}{2}$ $\text { or } \frac{2 \pi}{\mathrm{T}}=\frac{\pi}{2}$ $\mathrm{~T}=4 \mathrm{~s}$ The given time $(\mathrm{t})=3 \mathrm{~s}$ is $\frac{3 \mathrm{~T}}{4}$ At time $(\mathrm{t})=0$ particle is at $\mathrm{x}=\mathrm{A}$ (at extreme position) and at $\mathrm{t}=3 \mathrm{~s}=\frac{3 \mathrm{~T}}{4}$ it will be at mean position, $\mathrm{x}=0$. $\therefore$ Distance covered will be 3 r.
JCECE-2017
Oscillations
140374
A particle is executing simple harmonic motion. Its displacement to amplitude ratio when its kinetic energy is $84 \%$ of total energy is
1 $1: 16$
2 $2: 5$
3 $4: 25$
4 $21: 25$
Explanation:
B Let the total energy be, $\mathrm{E}_{\text {total }}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ And kinetic energy (K.E.) $=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ Now, it displacement to amplitude ratio is given as $\frac{K E}{E_{\text {total }}}=84 \%$ $\frac{K E}{E_{\text {total }}}=\frac{84}{100}$ $\frac{\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)}{\frac{1}{2} m \omega^{2} A^{2}}=\frac{84}{100}$ $\frac{\left(A^{2}-x^{2}\right)}{A^{2}}=\frac{84}{100}$ $\frac{1-\frac{x^{2}}{A^{2}}=\frac{84}{100}}{\frac{x^{2}}{A^{2}}=1-\frac{84}{100}=\frac{100-84}{100}}$ $\frac{x^{2}}{A^{2}}=\frac{16}{100}$ $\frac{x}{A}=\sqrt{\frac{16}{100}}$ $\frac{x}{A}=\frac{4}{10}=\frac{2}{5}$ $\frac{x}{A}=2: 5$ Hence, $\mathrm{x}: \mathrm{A}=2: 5$
140371
The displacement of a particle of mass $2 \mathrm{~g}$ executing $S H M$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here, $y$ is in metres and $t$ is in seconds. The kinetic energy of the particle, when $t=\frac{T}{4}$ is
1 $0.4 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $0.3 \mathrm{~J}$
Explanation:
D Displacement equation in S.H.M - $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$ $y=a \sin (\omega t+\phi)$ On comparing equation, we get - $\omega=4, \quad \mathrm{a}=5$ $\mathrm{~T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2} \text { second }$ $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\pi}{2 \times 4}=\frac{\pi}{8} \text { second }$ $\mathrm{y}=5 \sin \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ On differentiating both side wrt time $\frac{\mathrm{dy}}{\mathrm{dt}}=20 \cos \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ $\operatorname{Velocity}(v)=20 \cos \left(4 t+\frac{\pi}{3}\right)$ $\mathrm{v}=20 \cos \left(4 \times \frac{\pi}{8}+\frac{\pi}{3}\right)=20 \cos 150^{\circ}\left(\because \mathrm{t}=\frac{\pi}{8}\right)$ $\mathrm{v}=-20 \cos 30^{\circ}=\frac{-20 \sqrt{3}}{2} \mathrm{~m} / \mathrm{sec}$ Kinetic energy $($ K.E. $)=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \times \frac{2}{1000}(-10 \sqrt{3})^{2}$ $\text { K.E. }=\frac{1}{2} \times \frac{2}{1000} \times 100 \times 3=\frac{1 \times 300}{1000}$ $\text { K.E. }=0.3 \mathrm{~J}$
AP EAMCET (23.04.2018) Shift-2
Oscillations
140372
The potential energy of a simple harmonic oscillator of mass $2 \mathrm{~kg}$ at its mean position is 5 $\mathrm{J}$. If its total energy is $9 \mathrm{~J}$ and amplitude is 1 $\mathrm{cm}$, then its time period is
1 $\frac{\pi}{100} \mathrm{~s}$
2 $\frac{\pi}{50} \mathrm{~s}$
3 $\frac{\pi}{20} \mathrm{~s}$
4 $\frac{\pi}{10} \mathrm{~s}$
Explanation:
A Given, $\mathrm{m}=2 \mathrm{~kg}, \mathrm{a}=0.01 \mathrm{~m}$ Potential energy at mean position $=5 \mathrm{~J}$ Total energy $=9 \mathrm{~J}$ Total energy $=(\text { Potential energy })_{\text {mean }}+(\text { K.E })_{\max }$ $9=5+(\mathrm{K} . \mathrm{E})_{\max }$ $(\mathrm{K} . \mathrm{E})_{\max }=4 \mathrm{~J}$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{~J}$ $\mathrm{v}_{\text {max }}=\sqrt{\frac{4 \times 2}{\mathrm{~m}}}=\sqrt{\frac{4 \times 2}{2}}=2 \mathrm{~m} / \mathrm{sec}$ Maximum velocity $=\mathrm{a} \omega$ $2=0.01 \omega$ $\omega=200 \mathrm{rad} / \mathrm{sec}$ We know that $\mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \times \pi}{200}=\frac{\pi}{100} \mathrm{~s}$
AP EAMCET (22.04.2018) Shift-1
Oscillations
140373
A particle move according to the law $x=\operatorname{rcos} \frac{\pi t}{2}$. The distance covered by it in the time interval between $t=0$ and $t=3 \mathrm{~s}$ is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $3 \mathrm{r}$
4 $4 \mathrm{r}$
Explanation:
C Given that, $x=\operatorname{rcos} \frac{\pi t}{2}$ Comparing the given equation, we get - $\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}$ $\text { Then, } \omega=\frac{\pi}{2}$ $\text { or } \frac{2 \pi}{\mathrm{T}}=\frac{\pi}{2}$ $\mathrm{~T}=4 \mathrm{~s}$ The given time $(\mathrm{t})=3 \mathrm{~s}$ is $\frac{3 \mathrm{~T}}{4}$ At time $(\mathrm{t})=0$ particle is at $\mathrm{x}=\mathrm{A}$ (at extreme position) and at $\mathrm{t}=3 \mathrm{~s}=\frac{3 \mathrm{~T}}{4}$ it will be at mean position, $\mathrm{x}=0$. $\therefore$ Distance covered will be 3 r.
JCECE-2017
Oscillations
140374
A particle is executing simple harmonic motion. Its displacement to amplitude ratio when its kinetic energy is $84 \%$ of total energy is
1 $1: 16$
2 $2: 5$
3 $4: 25$
4 $21: 25$
Explanation:
B Let the total energy be, $\mathrm{E}_{\text {total }}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ And kinetic energy (K.E.) $=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ Now, it displacement to amplitude ratio is given as $\frac{K E}{E_{\text {total }}}=84 \%$ $\frac{K E}{E_{\text {total }}}=\frac{84}{100}$ $\frac{\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)}{\frac{1}{2} m \omega^{2} A^{2}}=\frac{84}{100}$ $\frac{\left(A^{2}-x^{2}\right)}{A^{2}}=\frac{84}{100}$ $\frac{1-\frac{x^{2}}{A^{2}}=\frac{84}{100}}{\frac{x^{2}}{A^{2}}=1-\frac{84}{100}=\frac{100-84}{100}}$ $\frac{x^{2}}{A^{2}}=\frac{16}{100}$ $\frac{x}{A}=\sqrt{\frac{16}{100}}$ $\frac{x}{A}=\frac{4}{10}=\frac{2}{5}$ $\frac{x}{A}=2: 5$ Hence, $\mathrm{x}: \mathrm{A}=2: 5$
140371
The displacement of a particle of mass $2 \mathrm{~g}$ executing $S H M$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here, $y$ is in metres and $t$ is in seconds. The kinetic energy of the particle, when $t=\frac{T}{4}$ is
1 $0.4 \mathrm{~J}$
2 $0.5 \mathrm{~J}$
3 $3 \mathrm{~J}$
4 $0.3 \mathrm{~J}$
Explanation:
D Displacement equation in S.H.M - $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$ $y=a \sin (\omega t+\phi)$ On comparing equation, we get - $\omega=4, \quad \mathrm{a}=5$ $\mathrm{~T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{\pi}{2} \text { second }$ $\mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\pi}{2 \times 4}=\frac{\pi}{8} \text { second }$ $\mathrm{y}=5 \sin \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ On differentiating both side wrt time $\frac{\mathrm{dy}}{\mathrm{dt}}=20 \cos \left(4 \mathrm{t}+\frac{\pi}{3}\right)$ $\operatorname{Velocity}(v)=20 \cos \left(4 t+\frac{\pi}{3}\right)$ $\mathrm{v}=20 \cos \left(4 \times \frac{\pi}{8}+\frac{\pi}{3}\right)=20 \cos 150^{\circ}\left(\because \mathrm{t}=\frac{\pi}{8}\right)$ $\mathrm{v}=-20 \cos 30^{\circ}=\frac{-20 \sqrt{3}}{2} \mathrm{~m} / \mathrm{sec}$ Kinetic energy $($ K.E. $)=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \times \frac{2}{1000}(-10 \sqrt{3})^{2}$ $\text { K.E. }=\frac{1}{2} \times \frac{2}{1000} \times 100 \times 3=\frac{1 \times 300}{1000}$ $\text { K.E. }=0.3 \mathrm{~J}$
AP EAMCET (23.04.2018) Shift-2
Oscillations
140372
The potential energy of a simple harmonic oscillator of mass $2 \mathrm{~kg}$ at its mean position is 5 $\mathrm{J}$. If its total energy is $9 \mathrm{~J}$ and amplitude is 1 $\mathrm{cm}$, then its time period is
1 $\frac{\pi}{100} \mathrm{~s}$
2 $\frac{\pi}{50} \mathrm{~s}$
3 $\frac{\pi}{20} \mathrm{~s}$
4 $\frac{\pi}{10} \mathrm{~s}$
Explanation:
A Given, $\mathrm{m}=2 \mathrm{~kg}, \mathrm{a}=0.01 \mathrm{~m}$ Potential energy at mean position $=5 \mathrm{~J}$ Total energy $=9 \mathrm{~J}$ Total energy $=(\text { Potential energy })_{\text {mean }}+(\text { K.E })_{\max }$ $9=5+(\mathrm{K} . \mathrm{E})_{\max }$ $(\mathrm{K} . \mathrm{E})_{\max }=4 \mathrm{~J}$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{~J}$ $\mathrm{v}_{\text {max }}=\sqrt{\frac{4 \times 2}{\mathrm{~m}}}=\sqrt{\frac{4 \times 2}{2}}=2 \mathrm{~m} / \mathrm{sec}$ Maximum velocity $=\mathrm{a} \omega$ $2=0.01 \omega$ $\omega=200 \mathrm{rad} / \mathrm{sec}$ We know that $\mathrm{T}=\frac{2 \pi}{\omega}$ $\mathrm{T}=\frac{2 \times \pi}{200}=\frac{\pi}{100} \mathrm{~s}$
AP EAMCET (22.04.2018) Shift-1
Oscillations
140373
A particle move according to the law $x=\operatorname{rcos} \frac{\pi t}{2}$. The distance covered by it in the time interval between $t=0$ and $t=3 \mathrm{~s}$ is
1 $\mathrm{r}$
2 $2 \mathrm{r}$
3 $3 \mathrm{r}$
4 $4 \mathrm{r}$
Explanation:
C Given that, $x=\operatorname{rcos} \frac{\pi t}{2}$ Comparing the given equation, we get - $\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}$ $\text { Then, } \omega=\frac{\pi}{2}$ $\text { or } \frac{2 \pi}{\mathrm{T}}=\frac{\pi}{2}$ $\mathrm{~T}=4 \mathrm{~s}$ The given time $(\mathrm{t})=3 \mathrm{~s}$ is $\frac{3 \mathrm{~T}}{4}$ At time $(\mathrm{t})=0$ particle is at $\mathrm{x}=\mathrm{A}$ (at extreme position) and at $\mathrm{t}=3 \mathrm{~s}=\frac{3 \mathrm{~T}}{4}$ it will be at mean position, $\mathrm{x}=0$. $\therefore$ Distance covered will be 3 r.
JCECE-2017
Oscillations
140374
A particle is executing simple harmonic motion. Its displacement to amplitude ratio when its kinetic energy is $84 \%$ of total energy is
1 $1: 16$
2 $2: 5$
3 $4: 25$
4 $21: 25$
Explanation:
B Let the total energy be, $\mathrm{E}_{\text {total }}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ And kinetic energy (K.E.) $=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ Now, it displacement to amplitude ratio is given as $\frac{K E}{E_{\text {total }}}=84 \%$ $\frac{K E}{E_{\text {total }}}=\frac{84}{100}$ $\frac{\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)}{\frac{1}{2} m \omega^{2} A^{2}}=\frac{84}{100}$ $\frac{\left(A^{2}-x^{2}\right)}{A^{2}}=\frac{84}{100}$ $\frac{1-\frac{x^{2}}{A^{2}}=\frac{84}{100}}{\frac{x^{2}}{A^{2}}=1-\frac{84}{100}=\frac{100-84}{100}}$ $\frac{x^{2}}{A^{2}}=\frac{16}{100}$ $\frac{x}{A}=\sqrt{\frac{16}{100}}$ $\frac{x}{A}=\frac{4}{10}=\frac{2}{5}$ $\frac{x}{A}=2: 5$ Hence, $\mathrm{x}: \mathrm{A}=2: 5$
