364120
A particle is executing SHM with an amplitude of \(4\;cm\). At what distance from equilibrium position its potential energy will be \(25 \%\) of its total energy?
1 \(\sqrt 2 \;cm\)
2 \(2\sqrt 3 \;cm\)
3 \(2\;cm\)
4 \(2\sqrt 2 \;cm\)
Explanation:
\(\dfrac{1}{2} m \omega^{2} x^{2}=\dfrac{1}{4} \times \dfrac{1}{2} m a^{2} \omega^{2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} x = \frac{a}{2} = \frac{4}{2}\;cm = 2\;cm\)
PHXI14:OSCILLATIONS
364121
In simple harmonic motion, the total mechanical energy of given system is \({E}\). If mass of oscillating particle \({P}\) is doubled, then the new energy of the system for same amplitude is
1 \({2 E}\)
2 \({E}\)
3 \({E \sqrt{2}}\)
4 \({E / \sqrt{2}}\)
Explanation:
Total mechanical energy of spring mass system is given by \(T . E .=\dfrac{1}{2} k A^{2}\) Where \({A=}\) Amplitude of oscillation It is independent of mass, so \(T.E.\) will remain same. So, correct option is (2).
JEE - 2024
PHXI14:OSCILLATIONS
364122
A particle undergoing SHM has the equation \(x = A\sin (\omega t + \phi )\), where \(x\) represents the displacement of the particle. The kinetic energy oscillates with time period
1 \(\dfrac{2 \pi}{\omega}\)
2 \(\dfrac{\pi}{\omega}\)
3 \(\dfrac{4 \pi}{\omega}\)
4 None
Explanation:
Frequency of oscillation of kinetic energy is doubled, i.e., \(2 \omega\) \(\therefore \mathrm{T}=\dfrac{2 \pi}{2 \omega}=\dfrac{\pi}{\omega}\)
PHXI14:OSCILLATIONS
364123
Which of the following quantities are always negative in a simple harmonic motion?
1 \(\vec{F} \cdot \vec{r}\).
2 \(\vec{v} \cdot \vec{r}\).
3 \(\vec{a} \cdot \vec{r}\)
4 Both (1) & (3)
Explanation:
Force and acceleration will be always towards the origin. Position vector defined from the origin is always away from the mean position. \(\bar{v}\) can be towards or away from the mean position. So \(\vec F.\vec r\,\,\& \,\,\vec a \cdot \vec r\) are (-) ve.
364120
A particle is executing SHM with an amplitude of \(4\;cm\). At what distance from equilibrium position its potential energy will be \(25 \%\) of its total energy?
1 \(\sqrt 2 \;cm\)
2 \(2\sqrt 3 \;cm\)
3 \(2\;cm\)
4 \(2\sqrt 2 \;cm\)
Explanation:
\(\dfrac{1}{2} m \omega^{2} x^{2}=\dfrac{1}{4} \times \dfrac{1}{2} m a^{2} \omega^{2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} x = \frac{a}{2} = \frac{4}{2}\;cm = 2\;cm\)
PHXI14:OSCILLATIONS
364121
In simple harmonic motion, the total mechanical energy of given system is \({E}\). If mass of oscillating particle \({P}\) is doubled, then the new energy of the system for same amplitude is
1 \({2 E}\)
2 \({E}\)
3 \({E \sqrt{2}}\)
4 \({E / \sqrt{2}}\)
Explanation:
Total mechanical energy of spring mass system is given by \(T . E .=\dfrac{1}{2} k A^{2}\) Where \({A=}\) Amplitude of oscillation It is independent of mass, so \(T.E.\) will remain same. So, correct option is (2).
JEE - 2024
PHXI14:OSCILLATIONS
364122
A particle undergoing SHM has the equation \(x = A\sin (\omega t + \phi )\), where \(x\) represents the displacement of the particle. The kinetic energy oscillates with time period
1 \(\dfrac{2 \pi}{\omega}\)
2 \(\dfrac{\pi}{\omega}\)
3 \(\dfrac{4 \pi}{\omega}\)
4 None
Explanation:
Frequency of oscillation of kinetic energy is doubled, i.e., \(2 \omega\) \(\therefore \mathrm{T}=\dfrac{2 \pi}{2 \omega}=\dfrac{\pi}{\omega}\)
PHXI14:OSCILLATIONS
364123
Which of the following quantities are always negative in a simple harmonic motion?
1 \(\vec{F} \cdot \vec{r}\).
2 \(\vec{v} \cdot \vec{r}\).
3 \(\vec{a} \cdot \vec{r}\)
4 Both (1) & (3)
Explanation:
Force and acceleration will be always towards the origin. Position vector defined from the origin is always away from the mean position. \(\bar{v}\) can be towards or away from the mean position. So \(\vec F.\vec r\,\,\& \,\,\vec a \cdot \vec r\) are (-) ve.
364120
A particle is executing SHM with an amplitude of \(4\;cm\). At what distance from equilibrium position its potential energy will be \(25 \%\) of its total energy?
1 \(\sqrt 2 \;cm\)
2 \(2\sqrt 3 \;cm\)
3 \(2\;cm\)
4 \(2\sqrt 2 \;cm\)
Explanation:
\(\dfrac{1}{2} m \omega^{2} x^{2}=\dfrac{1}{4} \times \dfrac{1}{2} m a^{2} \omega^{2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} x = \frac{a}{2} = \frac{4}{2}\;cm = 2\;cm\)
PHXI14:OSCILLATIONS
364121
In simple harmonic motion, the total mechanical energy of given system is \({E}\). If mass of oscillating particle \({P}\) is doubled, then the new energy of the system for same amplitude is
1 \({2 E}\)
2 \({E}\)
3 \({E \sqrt{2}}\)
4 \({E / \sqrt{2}}\)
Explanation:
Total mechanical energy of spring mass system is given by \(T . E .=\dfrac{1}{2} k A^{2}\) Where \({A=}\) Amplitude of oscillation It is independent of mass, so \(T.E.\) will remain same. So, correct option is (2).
JEE - 2024
PHXI14:OSCILLATIONS
364122
A particle undergoing SHM has the equation \(x = A\sin (\omega t + \phi )\), where \(x\) represents the displacement of the particle. The kinetic energy oscillates with time period
1 \(\dfrac{2 \pi}{\omega}\)
2 \(\dfrac{\pi}{\omega}\)
3 \(\dfrac{4 \pi}{\omega}\)
4 None
Explanation:
Frequency of oscillation of kinetic energy is doubled, i.e., \(2 \omega\) \(\therefore \mathrm{T}=\dfrac{2 \pi}{2 \omega}=\dfrac{\pi}{\omega}\)
PHXI14:OSCILLATIONS
364123
Which of the following quantities are always negative in a simple harmonic motion?
1 \(\vec{F} \cdot \vec{r}\).
2 \(\vec{v} \cdot \vec{r}\).
3 \(\vec{a} \cdot \vec{r}\)
4 Both (1) & (3)
Explanation:
Force and acceleration will be always towards the origin. Position vector defined from the origin is always away from the mean position. \(\bar{v}\) can be towards or away from the mean position. So \(\vec F.\vec r\,\,\& \,\,\vec a \cdot \vec r\) are (-) ve.
364120
A particle is executing SHM with an amplitude of \(4\;cm\). At what distance from equilibrium position its potential energy will be \(25 \%\) of its total energy?
1 \(\sqrt 2 \;cm\)
2 \(2\sqrt 3 \;cm\)
3 \(2\;cm\)
4 \(2\sqrt 2 \;cm\)
Explanation:
\(\dfrac{1}{2} m \omega^{2} x^{2}=\dfrac{1}{4} \times \dfrac{1}{2} m a^{2} \omega^{2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} x = \frac{a}{2} = \frac{4}{2}\;cm = 2\;cm\)
PHXI14:OSCILLATIONS
364121
In simple harmonic motion, the total mechanical energy of given system is \({E}\). If mass of oscillating particle \({P}\) is doubled, then the new energy of the system for same amplitude is
1 \({2 E}\)
2 \({E}\)
3 \({E \sqrt{2}}\)
4 \({E / \sqrt{2}}\)
Explanation:
Total mechanical energy of spring mass system is given by \(T . E .=\dfrac{1}{2} k A^{2}\) Where \({A=}\) Amplitude of oscillation It is independent of mass, so \(T.E.\) will remain same. So, correct option is (2).
JEE - 2024
PHXI14:OSCILLATIONS
364122
A particle undergoing SHM has the equation \(x = A\sin (\omega t + \phi )\), where \(x\) represents the displacement of the particle. The kinetic energy oscillates with time period
1 \(\dfrac{2 \pi}{\omega}\)
2 \(\dfrac{\pi}{\omega}\)
3 \(\dfrac{4 \pi}{\omega}\)
4 None
Explanation:
Frequency of oscillation of kinetic energy is doubled, i.e., \(2 \omega\) \(\therefore \mathrm{T}=\dfrac{2 \pi}{2 \omega}=\dfrac{\pi}{\omega}\)
PHXI14:OSCILLATIONS
364123
Which of the following quantities are always negative in a simple harmonic motion?
1 \(\vec{F} \cdot \vec{r}\).
2 \(\vec{v} \cdot \vec{r}\).
3 \(\vec{a} \cdot \vec{r}\)
4 Both (1) & (3)
Explanation:
Force and acceleration will be always towards the origin. Position vector defined from the origin is always away from the mean position. \(\bar{v}\) can be towards or away from the mean position. So \(\vec F.\vec r\,\,\& \,\,\vec a \cdot \vec r\) are (-) ve.
