Q. No.A particle executes simple harmonic motion with a time period \(T.\) It starts at its equilibrium position at \(t=0.\) How will the graph of its kinetic energy \((KE)\) versus time \((t)\) look like?
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Subtopic: Energy of SHM |
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A body of mass \(M\) and a charge \(q\) is connected to a spring of spring constant \(k.\) It oscillates along the \(x\text{-direction}\) about its equilibrium position, taken to be at \(x = 0,\) with an amplitude \(A.\) An electric field \(E\) is applied along the \(x\text{-direction}. \)
Which of the following statements is correct?
Which of the following statements is correct?
| 1. | The total energy of the system is \(\frac{1}{2}m\omega^2A^2+\frac{1}{2}\frac{q^2E^2}{k}.\) |
| 2. | The new equilibrium position is at a distance \(\frac{2qE} {k}\) from \(x = 0.\) |
| 3. | The new equilibrium position is at a distance \(\frac{qE} {2k}\) from \(x = 0.\) |
| 4. | The total energy of the system is \(\frac{1}{2}m\omega^2A^2-\frac{1}{2}\frac{q^2E^2}{k}.\) |
Subtopic: Energy of SHM |
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For a particle performing simple harmonic motion, the maximum potential energy is \(25~\text{J}\). What is the kinetic energy of the particle when it is at half of its amplitude?
1. \(18.75~\text{kJ}\)
2. \(18.75~\text{J}\)
3. \(9.45~\text{kJ}\)
4. \(9.45~\text{J}\)
1. \(18.75~\text{kJ}\)
2. \(18.75~\text{J}\)
3. \(9.45~\text{kJ}\)
4. \(9.45~\text{J}\)
Subtopic: Energy of SHM |
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For a particle undergoing linear simple harmonic motion (SHM), the graph showing the variation of kinetic energy, \(K\) with position, \(x\) of the particle is:
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Subtopic: Energy of SHM |
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A body is doing SHM with amplitude \(A.\) When it is at \(x = + {A \over 2}\), the ratio of kinetic energy to potential energy would be:
1. \(1:1\)
2. \(3:1\)
3. \(2:1\)
4. \(4:1\)
1. \(1:1\)
2. \(3:1\)
3. \(2:1\)
4. \(4:1\)
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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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. \(10\) kN/m
2. \(20\) kN/m
3. \(10000\) kN/m
4. \(20000\) kN/m
1. \(10\) kN/m
2. \(20\) kN/m
3. \(10000\) kN/m
4. \(20000\) kN/m
Subtopic: Energy of SHM |
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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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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 \(Y=\frac{A}{2}\) are:
| 1. | \( { K.E. }=\frac{k A^2}{8} \\ { P.E. }=\frac{3 k A^2}{8} \\ v=\frac{A}{3} \sqrt{\frac{k}{m}} \) | 2. | \({ K.E. }=\frac{3 k A^2}{8} \\ { P.E. }=\frac{k A^2}{8} \\ v=\frac{A}{2} \sqrt{\frac{3 k}{m}} \) |
| 3. | \({ K.E. }=\frac{3 k A^2}{8} \\ { P.E. }=\frac{k A^2}{4} \\ v=A \sqrt{\frac{3 k}{m}} \) | 4. | \({ K.E. }=\frac{k A^2}{4} \\ { P.E. }=\frac{3 k A^2}{8} \\ v=\frac{A}{4} \sqrt{\frac{3 k}{m}} \) |
Subtopic: Energy of SHM |
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A particle executing simple harmonic motion along \(x\text-\)axis, with amplitude \(A\) about the origin. Then the ratio of kinetic energy and total energy at \({x}={A\over 3}\) is:
1. \({8\over 9}\)
2. \({7\over 8}\)
3. \({1\over 9}\)
4. \({1\over 8}\)
1. \({8\over 9}\)
2. \({7\over 8}\)
3. \({1\over 9}\)
4. \({1\over 8}\)
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