NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI14:OSCILLATIONS
364099
The particle executing simple harmonic motion has a kinetic energy \(K_{0} \cos ^{2} \omega t\). The maximum values of the potential energy and the total energy are respectively:
1 0 and \(2 K_{0}\)
2 \(\dfrac{K_{0}}{2}\) and \(K_{0}\)
3 \(K_{0}\) and \(2 K_{0}\)
4 \(K_{0}\) and \(K_{0}\)
Explanation:
In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position. Now, \(E_{k}=K_{0} \cos ^{2} \omega t\) \(\therefore {\left( {{E_K}} \right)_{Max}} = {K_0}\) So, \(\left(E_{p}\right)_{\text {Max }}=K_{0}\) and \((E)_{\text {total }}=K_{0}\)
PHXI14:OSCILLATIONS
364100
The average energy in one time period in simple harmonic motion is
1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(m \omega^{2} a^{2}\)
3 \(1.5\,m{\omega ^2}{a^2}\)
4 None of these
Explanation:
Total energy in SHM is always constant and it is equal to \(\frac{1}{2}\,m\,{\omega ^2}{a^2}\). So \((T . E)_{\text {avg }}=\dfrac{1}{2} m \omega^{2} a^{2}\)
PHXI14:OSCILLATIONS
364101
The time period of the variation of potential energy of a particle executing SHM with period \(T\) is
1 \(T\)
2 \(T/4\)
3 \(T/2\)
4 \(2\;T\)
Explanation:
The potential energy of a particle executing SHM is periodic with time period \(\dfrac{T}{2}\).
PHXI14:OSCILLATIONS
364102
For a particle executing S.H.M. The displacement \(x\) is given by \(x=A \cos \omega t\). Identity the graph which represents the variation of potential energy (P.E.) as a function of time \(t\) and displacement \(x\)
1 II, IV
2 I, III
3 I, IV
4 II, III
Explanation:
At time \(t = 0,\,x = A,\) hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at \(x = 0,\) hence this is also correct.
364099
The particle executing simple harmonic motion has a kinetic energy \(K_{0} \cos ^{2} \omega t\). The maximum values of the potential energy and the total energy are respectively:
1 0 and \(2 K_{0}\)
2 \(\dfrac{K_{0}}{2}\) and \(K_{0}\)
3 \(K_{0}\) and \(2 K_{0}\)
4 \(K_{0}\) and \(K_{0}\)
Explanation:
In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position. Now, \(E_{k}=K_{0} \cos ^{2} \omega t\) \(\therefore {\left( {{E_K}} \right)_{Max}} = {K_0}\) So, \(\left(E_{p}\right)_{\text {Max }}=K_{0}\) and \((E)_{\text {total }}=K_{0}\)
PHXI14:OSCILLATIONS
364100
The average energy in one time period in simple harmonic motion is
1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(m \omega^{2} a^{2}\)
3 \(1.5\,m{\omega ^2}{a^2}\)
4 None of these
Explanation:
Total energy in SHM is always constant and it is equal to \(\frac{1}{2}\,m\,{\omega ^2}{a^2}\). So \((T . E)_{\text {avg }}=\dfrac{1}{2} m \omega^{2} a^{2}\)
PHXI14:OSCILLATIONS
364101
The time period of the variation of potential energy of a particle executing SHM with period \(T\) is
1 \(T\)
2 \(T/4\)
3 \(T/2\)
4 \(2\;T\)
Explanation:
The potential energy of a particle executing SHM is periodic with time period \(\dfrac{T}{2}\).
PHXI14:OSCILLATIONS
364102
For a particle executing S.H.M. The displacement \(x\) is given by \(x=A \cos \omega t\). Identity the graph which represents the variation of potential energy (P.E.) as a function of time \(t\) and displacement \(x\)
1 II, IV
2 I, III
3 I, IV
4 II, III
Explanation:
At time \(t = 0,\,x = A,\) hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at \(x = 0,\) hence this is also correct.
364099
The particle executing simple harmonic motion has a kinetic energy \(K_{0} \cos ^{2} \omega t\). The maximum values of the potential energy and the total energy are respectively:
1 0 and \(2 K_{0}\)
2 \(\dfrac{K_{0}}{2}\) and \(K_{0}\)
3 \(K_{0}\) and \(2 K_{0}\)
4 \(K_{0}\) and \(K_{0}\)
Explanation:
In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position. Now, \(E_{k}=K_{0} \cos ^{2} \omega t\) \(\therefore {\left( {{E_K}} \right)_{Max}} = {K_0}\) So, \(\left(E_{p}\right)_{\text {Max }}=K_{0}\) and \((E)_{\text {total }}=K_{0}\)
PHXI14:OSCILLATIONS
364100
The average energy in one time period in simple harmonic motion is
1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(m \omega^{2} a^{2}\)
3 \(1.5\,m{\omega ^2}{a^2}\)
4 None of these
Explanation:
Total energy in SHM is always constant and it is equal to \(\frac{1}{2}\,m\,{\omega ^2}{a^2}\). So \((T . E)_{\text {avg }}=\dfrac{1}{2} m \omega^{2} a^{2}\)
PHXI14:OSCILLATIONS
364101
The time period of the variation of potential energy of a particle executing SHM with period \(T\) is
1 \(T\)
2 \(T/4\)
3 \(T/2\)
4 \(2\;T\)
Explanation:
The potential energy of a particle executing SHM is periodic with time period \(\dfrac{T}{2}\).
PHXI14:OSCILLATIONS
364102
For a particle executing S.H.M. The displacement \(x\) is given by \(x=A \cos \omega t\). Identity the graph which represents the variation of potential energy (P.E.) as a function of time \(t\) and displacement \(x\)
1 II, IV
2 I, III
3 I, IV
4 II, III
Explanation:
At time \(t = 0,\,x = A,\) hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at \(x = 0,\) hence this is also correct.
364099
The particle executing simple harmonic motion has a kinetic energy \(K_{0} \cos ^{2} \omega t\). The maximum values of the potential energy and the total energy are respectively:
1 0 and \(2 K_{0}\)
2 \(\dfrac{K_{0}}{2}\) and \(K_{0}\)
3 \(K_{0}\) and \(2 K_{0}\)
4 \(K_{0}\) and \(K_{0}\)
Explanation:
In simple harmonic motion, the total energy of the particle is constant at all instants which is totally kinetic when particle is passing through the mean position and is totally potential when particle is passing through the extreme position. Now, \(E_{k}=K_{0} \cos ^{2} \omega t\) \(\therefore {\left( {{E_K}} \right)_{Max}} = {K_0}\) So, \(\left(E_{p}\right)_{\text {Max }}=K_{0}\) and \((E)_{\text {total }}=K_{0}\)
PHXI14:OSCILLATIONS
364100
The average energy in one time period in simple harmonic motion is
1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(m \omega^{2} a^{2}\)
3 \(1.5\,m{\omega ^2}{a^2}\)
4 None of these
Explanation:
Total energy in SHM is always constant and it is equal to \(\frac{1}{2}\,m\,{\omega ^2}{a^2}\). So \((T . E)_{\text {avg }}=\dfrac{1}{2} m \omega^{2} a^{2}\)
PHXI14:OSCILLATIONS
364101
The time period of the variation of potential energy of a particle executing SHM with period \(T\) is
1 \(T\)
2 \(T/4\)
3 \(T/2\)
4 \(2\;T\)
Explanation:
The potential energy of a particle executing SHM is periodic with time period \(\dfrac{T}{2}\).
PHXI14:OSCILLATIONS
364102
For a particle executing S.H.M. The displacement \(x\) is given by \(x=A \cos \omega t\). Identity the graph which represents the variation of potential energy (P.E.) as a function of time \(t\) and displacement \(x\)
1 II, IV
2 I, III
3 I, IV
4 II, III
Explanation:
At time \(t = 0,\,x = A,\) hence potential energy should be maximum. Therefore graph I is correct. Further in graph III. Potential energy is mimimum at \(x = 0,\) hence this is also correct.
