140389
A particle starting from mean position performs linear S.H.M. Its amplitude is ' $A$ ' and total energy is ' $E$ '. At what displacement its kinetic energy is $3 \mathrm{E} / 4$ ?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{3}$
3 $\frac{\mathrm{A}}{2}$
4 $\mathrm{A}$
Explanation:
C Kinetic energy of a particle in linear S.H.M is $\mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ And, Total energy $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ According to the quest $\frac{3}{4} \times \frac{1}{2} m \omega^{2} A^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $\frac{3}{4} A^{2}=\left(A^{2}-x^{2}\right)$ $x^{2}=A^{2}-\frac{3}{4} A^{2}$ $x^{2}=\frac{1}{4} A^{2}$ $x=\frac{A}{2}$
Oscillations
140390
A body oscillates simple harmonically with a period of 2 second, starting from the origin. After what time will its kinetic energy be $\mathbf{7 5 \%}$ of the total energy. $\left(\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$
1 $\frac{1}{6} \mathrm{~s}$
2 $\frac{1}{3} \mathrm{~s}$
3 $\frac{1}{12} \mathrm{~s}$
4 $\frac{1}{4} \mathrm{~s}$
Explanation:
A Given, $\mathrm{T}=2 \mathrm{~s}$ Kinetic energy of a body undergoing SHM is given by, $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \cos ^{2} \omega \mathrm{t}$ And $\quad(\mathrm{KE})_{\max }=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2}$ According to the question, $\mathrm{KE}=(\mathrm{KE})_{\max } \times \frac{75}{100}$ $\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \cos ^{2} \omega \mathrm{t}=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \times \frac{3}{4}$ $\cos \omega \mathrm{t}= \pm \frac{\sqrt{3}}{2}$ $\omega \mathrm{t}=\frac{\pi}{6}$ $\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$ $\mathrm{t}=\frac{\mathrm{T}}{12}=\frac{2}{12}=\frac{1}{6} \mathrm{~s}$ $\text { Or } \quad \frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$
AIEEE-2008
Oscillations
140391
The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$ (where, $\mathrm{E}$ is total energy)
Explanation:
A Let the potential energy of SHM be E then- $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Since, the particle is half way to its end point then- $\mathrm{y}=\frac{\mathrm{A}}{2}$ Hence, Potential Energy $(E)=\frac{1}{2} m \omega^{2}\left(\frac{A}{2}\right)^{2}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{m \omega}^{2} \mathrm{~A}^{2}\right)$ $=\frac{1}{4} \times \mathrm{E}$
MHT-CET-2007
Oscillations
140392
A particle is vibrating in a simple harmonic motion with an amplitude $4 \mathrm{~cm}$. At what displacement from the equilibrium is its energy half potential and half kinetic ?
1 $2 \sqrt{2} \mathrm{~cm}$
2 $\sqrt{2} \mathrm{~cm}$
3 $3 \mathrm{~cm}$
4 $1 \mathrm{~cm}$
Explanation:
A Given, amplitude (A) $=4 \mathrm{~cm}$ $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ $\text { P.E. }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ So, the total energy- $\text { T.E. }=\text { K.E. }+ \text { P.E. }$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}-\frac{1}{2} m \omega^{2} x^{2}+\frac{1}{2} m \omega^{2} x^{2}$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}$ According to question, P.E. $=$ K.E. becomes half Then, $\mathrm{x}^{2}=\mathrm{A}^{2}-\mathrm{x}^{2}$ $2 x^{2}=A^{2}$ $x^{2}=\frac{A^{2}}{2}$ $x=\frac{A}{\sqrt{2}}=\frac{4}{\sqrt{2}}$ $x=2 \sqrt{2} \mathrm{~cm}$
140389
A particle starting from mean position performs linear S.H.M. Its amplitude is ' $A$ ' and total energy is ' $E$ '. At what displacement its kinetic energy is $3 \mathrm{E} / 4$ ?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{3}$
3 $\frac{\mathrm{A}}{2}$
4 $\mathrm{A}$
Explanation:
C Kinetic energy of a particle in linear S.H.M is $\mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ And, Total energy $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ According to the quest $\frac{3}{4} \times \frac{1}{2} m \omega^{2} A^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $\frac{3}{4} A^{2}=\left(A^{2}-x^{2}\right)$ $x^{2}=A^{2}-\frac{3}{4} A^{2}$ $x^{2}=\frac{1}{4} A^{2}$ $x=\frac{A}{2}$
Oscillations
140390
A body oscillates simple harmonically with a period of 2 second, starting from the origin. After what time will its kinetic energy be $\mathbf{7 5 \%}$ of the total energy. $\left(\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$
1 $\frac{1}{6} \mathrm{~s}$
2 $\frac{1}{3} \mathrm{~s}$
3 $\frac{1}{12} \mathrm{~s}$
4 $\frac{1}{4} \mathrm{~s}$
Explanation:
A Given, $\mathrm{T}=2 \mathrm{~s}$ Kinetic energy of a body undergoing SHM is given by, $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \cos ^{2} \omega \mathrm{t}$ And $\quad(\mathrm{KE})_{\max }=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2}$ According to the question, $\mathrm{KE}=(\mathrm{KE})_{\max } \times \frac{75}{100}$ $\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \cos ^{2} \omega \mathrm{t}=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \times \frac{3}{4}$ $\cos \omega \mathrm{t}= \pm \frac{\sqrt{3}}{2}$ $\omega \mathrm{t}=\frac{\pi}{6}$ $\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$ $\mathrm{t}=\frac{\mathrm{T}}{12}=\frac{2}{12}=\frac{1}{6} \mathrm{~s}$ $\text { Or } \quad \frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$
AIEEE-2008
Oscillations
140391
The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$ (where, $\mathrm{E}$ is total energy)
Explanation:
A Let the potential energy of SHM be E then- $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Since, the particle is half way to its end point then- $\mathrm{y}=\frac{\mathrm{A}}{2}$ Hence, Potential Energy $(E)=\frac{1}{2} m \omega^{2}\left(\frac{A}{2}\right)^{2}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{m \omega}^{2} \mathrm{~A}^{2}\right)$ $=\frac{1}{4} \times \mathrm{E}$
MHT-CET-2007
Oscillations
140392
A particle is vibrating in a simple harmonic motion with an amplitude $4 \mathrm{~cm}$. At what displacement from the equilibrium is its energy half potential and half kinetic ?
1 $2 \sqrt{2} \mathrm{~cm}$
2 $\sqrt{2} \mathrm{~cm}$
3 $3 \mathrm{~cm}$
4 $1 \mathrm{~cm}$
Explanation:
A Given, amplitude (A) $=4 \mathrm{~cm}$ $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ $\text { P.E. }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ So, the total energy- $\text { T.E. }=\text { K.E. }+ \text { P.E. }$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}-\frac{1}{2} m \omega^{2} x^{2}+\frac{1}{2} m \omega^{2} x^{2}$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}$ According to question, P.E. $=$ K.E. becomes half Then, $\mathrm{x}^{2}=\mathrm{A}^{2}-\mathrm{x}^{2}$ $2 x^{2}=A^{2}$ $x^{2}=\frac{A^{2}}{2}$ $x=\frac{A}{\sqrt{2}}=\frac{4}{\sqrt{2}}$ $x=2 \sqrt{2} \mathrm{~cm}$
140389
A particle starting from mean position performs linear S.H.M. Its amplitude is ' $A$ ' and total energy is ' $E$ '. At what displacement its kinetic energy is $3 \mathrm{E} / 4$ ?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{3}$
3 $\frac{\mathrm{A}}{2}$
4 $\mathrm{A}$
Explanation:
C Kinetic energy of a particle in linear S.H.M is $\mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ And, Total energy $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ According to the quest $\frac{3}{4} \times \frac{1}{2} m \omega^{2} A^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $\frac{3}{4} A^{2}=\left(A^{2}-x^{2}\right)$ $x^{2}=A^{2}-\frac{3}{4} A^{2}$ $x^{2}=\frac{1}{4} A^{2}$ $x=\frac{A}{2}$
Oscillations
140390
A body oscillates simple harmonically with a period of 2 second, starting from the origin. After what time will its kinetic energy be $\mathbf{7 5 \%}$ of the total energy. $\left(\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$
1 $\frac{1}{6} \mathrm{~s}$
2 $\frac{1}{3} \mathrm{~s}$
3 $\frac{1}{12} \mathrm{~s}$
4 $\frac{1}{4} \mathrm{~s}$
Explanation:
A Given, $\mathrm{T}=2 \mathrm{~s}$ Kinetic energy of a body undergoing SHM is given by, $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \cos ^{2} \omega \mathrm{t}$ And $\quad(\mathrm{KE})_{\max }=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2}$ According to the question, $\mathrm{KE}=(\mathrm{KE})_{\max } \times \frac{75}{100}$ $\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \cos ^{2} \omega \mathrm{t}=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \times \frac{3}{4}$ $\cos \omega \mathrm{t}= \pm \frac{\sqrt{3}}{2}$ $\omega \mathrm{t}=\frac{\pi}{6}$ $\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$ $\mathrm{t}=\frac{\mathrm{T}}{12}=\frac{2}{12}=\frac{1}{6} \mathrm{~s}$ $\text { Or } \quad \frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$
AIEEE-2008
Oscillations
140391
The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$ (where, $\mathrm{E}$ is total energy)
Explanation:
A Let the potential energy of SHM be E then- $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Since, the particle is half way to its end point then- $\mathrm{y}=\frac{\mathrm{A}}{2}$ Hence, Potential Energy $(E)=\frac{1}{2} m \omega^{2}\left(\frac{A}{2}\right)^{2}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{m \omega}^{2} \mathrm{~A}^{2}\right)$ $=\frac{1}{4} \times \mathrm{E}$
MHT-CET-2007
Oscillations
140392
A particle is vibrating in a simple harmonic motion with an amplitude $4 \mathrm{~cm}$. At what displacement from the equilibrium is its energy half potential and half kinetic ?
1 $2 \sqrt{2} \mathrm{~cm}$
2 $\sqrt{2} \mathrm{~cm}$
3 $3 \mathrm{~cm}$
4 $1 \mathrm{~cm}$
Explanation:
A Given, amplitude (A) $=4 \mathrm{~cm}$ $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ $\text { P.E. }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ So, the total energy- $\text { T.E. }=\text { K.E. }+ \text { P.E. }$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}-\frac{1}{2} m \omega^{2} x^{2}+\frac{1}{2} m \omega^{2} x^{2}$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}$ According to question, P.E. $=$ K.E. becomes half Then, $\mathrm{x}^{2}=\mathrm{A}^{2}-\mathrm{x}^{2}$ $2 x^{2}=A^{2}$ $x^{2}=\frac{A^{2}}{2}$ $x=\frac{A}{\sqrt{2}}=\frac{4}{\sqrt{2}}$ $x=2 \sqrt{2} \mathrm{~cm}$
140389
A particle starting from mean position performs linear S.H.M. Its amplitude is ' $A$ ' and total energy is ' $E$ '. At what displacement its kinetic energy is $3 \mathrm{E} / 4$ ?
1 $\frac{\mathrm{A}}{4}$
2 $\frac{\mathrm{A}}{3}$
3 $\frac{\mathrm{A}}{2}$
4 $\mathrm{A}$
Explanation:
C Kinetic energy of a particle in linear S.H.M is $\mathrm{K}=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ And, Total energy $=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}$ According to the quest $\frac{3}{4} \times \frac{1}{2} m \omega^{2} A^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$ $\frac{3}{4} A^{2}=\left(A^{2}-x^{2}\right)$ $x^{2}=A^{2}-\frac{3}{4} A^{2}$ $x^{2}=\frac{1}{4} A^{2}$ $x=\frac{A}{2}$
Oscillations
140390
A body oscillates simple harmonically with a period of 2 second, starting from the origin. After what time will its kinetic energy be $\mathbf{7 5 \%}$ of the total energy. $\left(\sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$
1 $\frac{1}{6} \mathrm{~s}$
2 $\frac{1}{3} \mathrm{~s}$
3 $\frac{1}{12} \mathrm{~s}$
4 $\frac{1}{4} \mathrm{~s}$
Explanation:
A Given, $\mathrm{T}=2 \mathrm{~s}$ Kinetic energy of a body undergoing SHM is given by, $\mathrm{KE}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \cos ^{2} \omega \mathrm{t}$ And $\quad(\mathrm{KE})_{\max }=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2}$ According to the question, $\mathrm{KE}=(\mathrm{KE})_{\max } \times \frac{75}{100}$ $\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \cos ^{2} \omega \mathrm{t}=\frac{\mathrm{m} \omega^{2} \mathrm{~A}^{2}}{2} \times \frac{3}{4}$ $\cos \omega \mathrm{t}= \pm \frac{\sqrt{3}}{2}$ $\omega \mathrm{t}=\frac{\pi}{6}$ $\frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$ $\mathrm{t}=\frac{\mathrm{T}}{12}=\frac{2}{12}=\frac{1}{6} \mathrm{~s}$ $\text { Or } \quad \frac{2 \pi}{\mathrm{T}} \times \mathrm{t}=\frac{\pi}{6}$
AIEEE-2008
Oscillations
140391
The potential energy of a simple harmonic oscillator, when the particle is half way to its end point is
1 $\frac{1}{4} \mathrm{E}$
2 $\frac{1}{2} \mathrm{E}$
3 $\frac{2}{3} \mathrm{E}$
4 $\frac{1}{8} \mathrm{E}$ (where, $\mathrm{E}$ is total energy)
Explanation:
A Let the potential energy of SHM be E then- $\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{y}^{2}$ Since, the particle is half way to its end point then- $\mathrm{y}=\frac{\mathrm{A}}{2}$ Hence, Potential Energy $(E)=\frac{1}{2} m \omega^{2}\left(\frac{A}{2}\right)^{2}$ $=\frac{1}{4}\left(\frac{1}{2} \mathrm{m \omega}^{2} \mathrm{~A}^{2}\right)$ $=\frac{1}{4} \times \mathrm{E}$
MHT-CET-2007
Oscillations
140392
A particle is vibrating in a simple harmonic motion with an amplitude $4 \mathrm{~cm}$. At what displacement from the equilibrium is its energy half potential and half kinetic ?
1 $2 \sqrt{2} \mathrm{~cm}$
2 $\sqrt{2} \mathrm{~cm}$
3 $3 \mathrm{~cm}$
4 $1 \mathrm{~cm}$
Explanation:
A Given, amplitude (A) $=4 \mathrm{~cm}$ $\text { K.E. }=\frac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$ $\text { P.E. }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{x}^{2}$ So, the total energy- $\text { T.E. }=\text { K.E. }+ \text { P.E. }$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}-\frac{1}{2} m \omega^{2} x^{2}+\frac{1}{2} m \omega^{2} x^{2}$ $\text { T.E. }=\frac{1}{2} m \omega^{2} A^{2}$ According to question, P.E. $=$ K.E. becomes half Then, $\mathrm{x}^{2}=\mathrm{A}^{2}-\mathrm{x}^{2}$ $2 x^{2}=A^{2}$ $x^{2}=\frac{A^{2}}{2}$ $x=\frac{A}{\sqrt{2}}=\frac{4}{\sqrt{2}}$ $x=2 \sqrt{2} \mathrm{~cm}$
