Q. No.The kinetic energy of a particle executing simple harmonic motion varies periodically with an angular frequency of \(176~\text{rad/s}.\) The frequency (in Hz) of the oscillator is: \(\left ( \pi =22/7 \right )\)
1. \(14\)
2. \(88\)
3. \(28\)
4. \(176\)
1. \(14\)
2. \(88\)
3. \(28\)
4. \(176\)
Subtopic: Energy of SHM |
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The displacement of a particle, executing simple harmonic motion with time period \(T\), is expressed as \(x(t)={A} \sin \omega {t},\) where \(A\) is the amplitude. The maximum value of potential energy of the oscillator is found at \(t=\dfrac{T}{2\beta}\). The value of \(\beta\) is:
1. \(4\)
2. \(2\)
3. \(5\)
4. \(1\)
1. \(4\)
2. \(2\)
3. \(5\)
4. \(1\)
Subtopic: Energy of SHM |
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A particle oscillates along the \(x\text-\)axis according to the law, \(x(t)=x_0 \sin^2\left(\dfrac{t}{2}\right) \) where \(x_0=1 .\) The kinetic energy \(K\) of the particle as a function of \(x\) is correctly represented by the graph:
| 1. |  |
| 2. |  |
| 3. |  |
| 4. |  |
Subtopic: Energy of SHM |
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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. \(\mathrm{E} \sqrt{2}\)
2. \(E\)
3. \(2E\)
4. \(\frac{E}{\sqrt{2}}\)
1. \(\mathrm{E} \sqrt{2}\)
2. \(E\)
3. \(2E\)
4. \(\frac{E}{\sqrt{2}}\)
Subtopic: Energy of SHM |
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A particle executes simple harmonic motion (SHM) along the \(x\text-\)axis with amplitude \(A\) about the origin. What is the ratio of the particle's kinetic energy to its total energy when its displacement is \(x=\dfrac{A}{3} \text{?}\)
| 1. | \(\dfrac{8}{9}\) | 2. | \(\dfrac{7}{8}\) |
| 3. | \(\dfrac{1}{9}\) | 4. | \(\dfrac{1}{8}\) |
Subtopic: Energy of SHM |
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A particle performing simple harmonic motion according to \(y = A \sin\omega t\). Then its kinetic energy \((K.E.),\) potential energy \((P.E.),\) and speed \((v)\) at the position \(Y=\dfrac{A}{2}\) are:
| 1. | \( { K.E. }=\dfrac{k A^2}{8} \\ { P.E. }=\dfrac{3 k A^2}{8} \\ v=\dfrac{A}{3} \sqrt{\dfrac{k}{m}} \) | 2. | \({ K.E. }=\dfrac{3 k A^2}{8} \\ { P.E. }=\dfrac{k A^2}{8} \\ v=\dfrac{A}{2} \sqrt{\dfrac{3 k}{m}} \) |
| 3. | \({ K.E. }=\dfrac{3 k A^2}{8} \\ { P.E. }=\dfrac{k A^2}{4} \\ v=A \sqrt{\dfrac{3 k}{m}} \) | 4. | \({ K.E. }=\dfrac{k A^2}{4} \\ { P.E. }=\dfrac{3 k A^2}{8} \\ v=\dfrac{A}{4} \sqrt{\dfrac{3 k}{m}} \) |
Subtopic: Energy of SHM |
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At a certain instant, kinetic energy and potential energy for a particle executing SHM are \(0.4 \mathrm{~J} \) and \(0.5 \mathrm{~J}\) respectively. Then the amplitude of SHM, if the frequency of oscillations is \(\frac{25}{\pi} ~\text{Hz}\) and the mass of the particle is \( 0.2 ~\text{kg},\) is:
1. \(5~\text{cm}\)
2. \(3~\text{cm}\)
3. \(4~\text{cm}\)
4. \(6~\text{cm}\)
1. \(5~\text{cm}\)
2. \(3~\text{cm}\)
3. \(4~\text{cm}\)
4. \(6~\text{cm}\)
Subtopic: Energy of SHM |
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A particle is performing SHM having position \( x = A~\text {cos} (30^\circ) , \) and \(A = 40 ~\text {cm} \). If its kinetic energy at this position is \(200\) J, then the value of force constant is:
1. \(10~\text{kN/m}\)
2. \(20~\text{kN/m}\)
3. \(10000~\text{kN/m}\)
4. \(20000~\text{kN/m}\)
1. \(10~\text{kN/m}\)
2. \(20~\text{kN/m}\)
3. \(10000~\text{kN/m}\)
4. \(20000~\text{kN/m}\)
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A particle is performing simple harmonic motion (SHM), whose distance from the mean position varies as \(x = A~ \mathrm{sin} \omega t.\) What would be the position of the particle from the mean position where kinetic energy and potential energy are equal?
| 1. | \(\left({{{A}\over{2}}}\right)\) | 2. | \(\left({{{A}\over{\sqrt{2}}}}\right)\) |
| 3. | \(\left({{{A}\over{2\sqrt{2}}}}\right)\) | 4. | \(\left({{{A}\over{4}}}\right)\) |
Subtopic: Energy of SHM |
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Which graph correctly represents the difference \((d)\) between the total energy and the potential energy of a particle in linear simple harmonic motion (SHM) as a function of its position \(x,\) where \(x=0\) denotes the mean position?
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2. |  |
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4. |  |
Subtopic: Energy of SHM |
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