Force, Energy and their relation in Simple Harmonic Motion
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PHXI14:OSCILLATIONS

364116 A particle executes SHM along a straight line. It starts its motion at extreme position at \(t = 0\). The time interval after which its kinetic energy will be equal to its potential energy for the first time is \((T = \) Time period)

1 \(T/5\)
2 \(T / 4\)
3 \(T / 6\)
4 \(T/8\)
PHXI14:OSCILLATIONS

364117 The maximum potential energy of a block executing simple harmonic motion is \(25\,J.A\) is amplitude of oscillation. At \(\dfrac{A}{2}\), the kinetic energy of the block is

1 \(18.75\;J\)
2 \(12.5\;J\)
3 \(37.5\;J\)
4 \(9.75\;J\)
JEE - 2023
PHXI14:OSCILLATIONS

364118 The P.E. of a particle executing SHM at a distance \(x\) from its equilibrium position is

1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(\dfrac{1}{2} m \omega^{2} x^{2}\)
3 Zero
4 \(\dfrac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)\)
PHXI14:OSCILLATIONS

364119 When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is:

1 \(a / 4\)
2 \(a / 3\)
3 \(a / 2\)
4 \(2 a / 3\)
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PHXI14:OSCILLATIONS

364116 A particle executes SHM along a straight line. It starts its motion at extreme position at \(t = 0\). The time interval after which its kinetic energy will be equal to its potential energy for the first time is \((T = \) Time period)

1 \(T/5\)
2 \(T / 4\)
3 \(T / 6\)
4 \(T/8\)
PHXI14:OSCILLATIONS

364117 The maximum potential energy of a block executing simple harmonic motion is \(25\,J.A\) is amplitude of oscillation. At \(\dfrac{A}{2}\), the kinetic energy of the block is

1 \(18.75\;J\)
2 \(12.5\;J\)
3 \(37.5\;J\)
4 \(9.75\;J\)
JEE - 2023
PHXI14:OSCILLATIONS

364118 The P.E. of a particle executing SHM at a distance \(x\) from its equilibrium position is

1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(\dfrac{1}{2} m \omega^{2} x^{2}\)
3 Zero
4 \(\dfrac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)\)
PHXI14:OSCILLATIONS

364119 When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is:

1 \(a / 4\)
2 \(a / 3\)
3 \(a / 2\)
4 \(2 a / 3\)
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PHXI14:OSCILLATIONS

364116 A particle executes SHM along a straight line. It starts its motion at extreme position at \(t = 0\). The time interval after which its kinetic energy will be equal to its potential energy for the first time is \((T = \) Time period)

1 \(T/5\)
2 \(T / 4\)
3 \(T / 6\)
4 \(T/8\)
PHXI14:OSCILLATIONS

364117 The maximum potential energy of a block executing simple harmonic motion is \(25\,J.A\) is amplitude of oscillation. At \(\dfrac{A}{2}\), the kinetic energy of the block is

1 \(18.75\;J\)
2 \(12.5\;J\)
3 \(37.5\;J\)
4 \(9.75\;J\)
JEE - 2023
PHXI14:OSCILLATIONS

364118 The P.E. of a particle executing SHM at a distance \(x\) from its equilibrium position is

1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(\dfrac{1}{2} m \omega^{2} x^{2}\)
3 Zero
4 \(\dfrac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)\)
PHXI14:OSCILLATIONS

364119 When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is:

1 \(a / 4\)
2 \(a / 3\)
3 \(a / 2\)
4 \(2 a / 3\)
Join With Us for More Updates
PHXI14:OSCILLATIONS

364116 A particle executes SHM along a straight line. It starts its motion at extreme position at \(t = 0\). The time interval after which its kinetic energy will be equal to its potential energy for the first time is \((T = \) Time period)

1 \(T/5\)
2 \(T / 4\)
3 \(T / 6\)
4 \(T/8\)
PHXI14:OSCILLATIONS

364117 The maximum potential energy of a block executing simple harmonic motion is \(25\,J.A\) is amplitude of oscillation. At \(\dfrac{A}{2}\), the kinetic energy of the block is

1 \(18.75\;J\)
2 \(12.5\;J\)
3 \(37.5\;J\)
4 \(9.75\;J\)
JEE - 2023
PHXI14:OSCILLATIONS

364118 The P.E. of a particle executing SHM at a distance \(x\) from its equilibrium position is

1 \(\dfrac{1}{2} m \omega^{2} a^{2}\)
2 \(\dfrac{1}{2} m \omega^{2} x^{2}\)
3 Zero
4 \(\dfrac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)\)
PHXI14:OSCILLATIONS

364119 When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is:

1 \(a / 4\)
2 \(a / 3\)
3 \(a / 2\)
4 \(2 a / 3\)
For More Updates join with Us