364103
The angular velocity and the amplitude of a simple pendulum is \(\omega\) and A respectively. At a displacement \(x\) from the mean position if its kinetic energy is \(T\) and potential energy is \(V\), then the ratio of \(T\,\,{\rm{to}}\,\,V\) is
Potential energy, \(V=\dfrac{1}{2} m \omega^{2} x^{2}\) and kinetic energy \(T=\dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)\) \(\therefore \dfrac{T}{V}=\dfrac{\mathrm{A}^{2}-x^{2}}{x^{2}}\)
PHXI14:OSCILLATIONS
364104
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where \(E\) is the total energy)
1 \(\dfrac{2}{3} E\)
2 \(\dfrac{1}{8} E\)
3 \(\dfrac{1}{4} E\)
4 \(\dfrac{1}{2} E\)
Explanation:
\(\dfrac{U}{E}=\dfrac{\dfrac{1}{2} K y^{2}}{\dfrac{1}{2} K a^{2}}=\dfrac{y^{2}}{a^{2}}=\dfrac{\left(\dfrac{a}{2}\right)^{2}}{a^{2}}=\dfrac{1}{4} \Rightarrow U=\dfrac{E}{4}\).
PHXI14:OSCILLATIONS
364105
The potential energy of a particle executing S.H.M. is \(2.5\,J\), when its displacement is half of amplitude. The total energy of the particle be
1 \(18\,J\)
2 \(10\,J\)
3 \(12\,J\)
4 \(2.5\,J\)
Explanation:
\(\dfrac{\text { Potential energy }(U)}{\text { Total energy }(E)}=\dfrac{\dfrac{1}{2} m \omega^{2} y^{2}}{\dfrac{1}{2} m \omega^{2} a^{2}}=\dfrac{y^{2}}{a^{2}}\) \({\rm{ }}So{\rm{ }}\frac{{2.5}}{E} = \frac{{{{\left( {\frac{a}{2}} \right)}^2}}}{{{a^2}}} \Rightarrow E = 10\;J\)
PHXI14:OSCILLATIONS
364106
The K. E and P. E of a particle executing SHM with amplitude \(A\) will be equal when its displacement is
1 \(A/2\)
2 \(A\sqrt 2 \)
3 \(A\sqrt {2/3} \)
4 \(A/\sqrt 2 \)
Explanation:
\(\mathrm{KE}=\mathrm{PE}\) \(\dfrac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-y^{2}\right)=\dfrac{1}{2} m \omega^{2} y^{2} \Rightarrow y=\dfrac{A}{\sqrt{2}}\)
364103
The angular velocity and the amplitude of a simple pendulum is \(\omega\) and A respectively. At a displacement \(x\) from the mean position if its kinetic energy is \(T\) and potential energy is \(V\), then the ratio of \(T\,\,{\rm{to}}\,\,V\) is
Potential energy, \(V=\dfrac{1}{2} m \omega^{2} x^{2}\) and kinetic energy \(T=\dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)\) \(\therefore \dfrac{T}{V}=\dfrac{\mathrm{A}^{2}-x^{2}}{x^{2}}\)
PHXI14:OSCILLATIONS
364104
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where \(E\) is the total energy)
1 \(\dfrac{2}{3} E\)
2 \(\dfrac{1}{8} E\)
3 \(\dfrac{1}{4} E\)
4 \(\dfrac{1}{2} E\)
Explanation:
\(\dfrac{U}{E}=\dfrac{\dfrac{1}{2} K y^{2}}{\dfrac{1}{2} K a^{2}}=\dfrac{y^{2}}{a^{2}}=\dfrac{\left(\dfrac{a}{2}\right)^{2}}{a^{2}}=\dfrac{1}{4} \Rightarrow U=\dfrac{E}{4}\).
PHXI14:OSCILLATIONS
364105
The potential energy of a particle executing S.H.M. is \(2.5\,J\), when its displacement is half of amplitude. The total energy of the particle be
1 \(18\,J\)
2 \(10\,J\)
3 \(12\,J\)
4 \(2.5\,J\)
Explanation:
\(\dfrac{\text { Potential energy }(U)}{\text { Total energy }(E)}=\dfrac{\dfrac{1}{2} m \omega^{2} y^{2}}{\dfrac{1}{2} m \omega^{2} a^{2}}=\dfrac{y^{2}}{a^{2}}\) \({\rm{ }}So{\rm{ }}\frac{{2.5}}{E} = \frac{{{{\left( {\frac{a}{2}} \right)}^2}}}{{{a^2}}} \Rightarrow E = 10\;J\)
PHXI14:OSCILLATIONS
364106
The K. E and P. E of a particle executing SHM with amplitude \(A\) will be equal when its displacement is
1 \(A/2\)
2 \(A\sqrt 2 \)
3 \(A\sqrt {2/3} \)
4 \(A/\sqrt 2 \)
Explanation:
\(\mathrm{KE}=\mathrm{PE}\) \(\dfrac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-y^{2}\right)=\dfrac{1}{2} m \omega^{2} y^{2} \Rightarrow y=\dfrac{A}{\sqrt{2}}\)
364103
The angular velocity and the amplitude of a simple pendulum is \(\omega\) and A respectively. At a displacement \(x\) from the mean position if its kinetic energy is \(T\) and potential energy is \(V\), then the ratio of \(T\,\,{\rm{to}}\,\,V\) is
Potential energy, \(V=\dfrac{1}{2} m \omega^{2} x^{2}\) and kinetic energy \(T=\dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)\) \(\therefore \dfrac{T}{V}=\dfrac{\mathrm{A}^{2}-x^{2}}{x^{2}}\)
PHXI14:OSCILLATIONS
364104
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where \(E\) is the total energy)
1 \(\dfrac{2}{3} E\)
2 \(\dfrac{1}{8} E\)
3 \(\dfrac{1}{4} E\)
4 \(\dfrac{1}{2} E\)
Explanation:
\(\dfrac{U}{E}=\dfrac{\dfrac{1}{2} K y^{2}}{\dfrac{1}{2} K a^{2}}=\dfrac{y^{2}}{a^{2}}=\dfrac{\left(\dfrac{a}{2}\right)^{2}}{a^{2}}=\dfrac{1}{4} \Rightarrow U=\dfrac{E}{4}\).
PHXI14:OSCILLATIONS
364105
The potential energy of a particle executing S.H.M. is \(2.5\,J\), when its displacement is half of amplitude. The total energy of the particle be
1 \(18\,J\)
2 \(10\,J\)
3 \(12\,J\)
4 \(2.5\,J\)
Explanation:
\(\dfrac{\text { Potential energy }(U)}{\text { Total energy }(E)}=\dfrac{\dfrac{1}{2} m \omega^{2} y^{2}}{\dfrac{1}{2} m \omega^{2} a^{2}}=\dfrac{y^{2}}{a^{2}}\) \({\rm{ }}So{\rm{ }}\frac{{2.5}}{E} = \frac{{{{\left( {\frac{a}{2}} \right)}^2}}}{{{a^2}}} \Rightarrow E = 10\;J\)
PHXI14:OSCILLATIONS
364106
The K. E and P. E of a particle executing SHM with amplitude \(A\) will be equal when its displacement is
1 \(A/2\)
2 \(A\sqrt 2 \)
3 \(A\sqrt {2/3} \)
4 \(A/\sqrt 2 \)
Explanation:
\(\mathrm{KE}=\mathrm{PE}\) \(\dfrac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-y^{2}\right)=\dfrac{1}{2} m \omega^{2} y^{2} \Rightarrow y=\dfrac{A}{\sqrt{2}}\)
364103
The angular velocity and the amplitude of a simple pendulum is \(\omega\) and A respectively. At a displacement \(x\) from the mean position if its kinetic energy is \(T\) and potential energy is \(V\), then the ratio of \(T\,\,{\rm{to}}\,\,V\) is
Potential energy, \(V=\dfrac{1}{2} m \omega^{2} x^{2}\) and kinetic energy \(T=\dfrac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)\) \(\therefore \dfrac{T}{V}=\dfrac{\mathrm{A}^{2}-x^{2}}{x^{2}}\)
PHXI14:OSCILLATIONS
364104
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where \(E\) is the total energy)
1 \(\dfrac{2}{3} E\)
2 \(\dfrac{1}{8} E\)
3 \(\dfrac{1}{4} E\)
4 \(\dfrac{1}{2} E\)
Explanation:
\(\dfrac{U}{E}=\dfrac{\dfrac{1}{2} K y^{2}}{\dfrac{1}{2} K a^{2}}=\dfrac{y^{2}}{a^{2}}=\dfrac{\left(\dfrac{a}{2}\right)^{2}}{a^{2}}=\dfrac{1}{4} \Rightarrow U=\dfrac{E}{4}\).
PHXI14:OSCILLATIONS
364105
The potential energy of a particle executing S.H.M. is \(2.5\,J\), when its displacement is half of amplitude. The total energy of the particle be
1 \(18\,J\)
2 \(10\,J\)
3 \(12\,J\)
4 \(2.5\,J\)
Explanation:
\(\dfrac{\text { Potential energy }(U)}{\text { Total energy }(E)}=\dfrac{\dfrac{1}{2} m \omega^{2} y^{2}}{\dfrac{1}{2} m \omega^{2} a^{2}}=\dfrac{y^{2}}{a^{2}}\) \({\rm{ }}So{\rm{ }}\frac{{2.5}}{E} = \frac{{{{\left( {\frac{a}{2}} \right)}^2}}}{{{a^2}}} \Rightarrow E = 10\;J\)
PHXI14:OSCILLATIONS
364106
The K. E and P. E of a particle executing SHM with amplitude \(A\) will be equal when its displacement is
1 \(A/2\)
2 \(A\sqrt 2 \)
3 \(A\sqrt {2/3} \)
4 \(A/\sqrt 2 \)
Explanation:
\(\mathrm{KE}=\mathrm{PE}\) \(\dfrac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-y^{2}\right)=\dfrac{1}{2} m \omega^{2} y^{2} \Rightarrow y=\dfrac{A}{\sqrt{2}}\)
