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Chapter 11: Problem 22

The total energy of the body executing S.H.M. is \(E\). Then the kinetic energy when the displacement is half of the amplitude is (a) \(E / 2\) (b) \(E / 4\) (c) \(3 E / 4\) (d) \(\sqrt{3} E / 4\)

Short Answer

Expert verified
Hence, if the displacement is half of the amplitude in Simple Harmonic Motion, the kinetic energy is \(\frac{3}{4} E\). So, the answer is (c) \(\frac{3}{4} E\)

Step by step solution

01

Define the displacement and kinetic energy

In Simple Harmonic Motion, if \(x = \frac{A}{2}\) corresponds to half the amplitude and \(K\) is the kinetic energy, then the kinetic energy \(K = E - U\), where \(U\) is the potential energy.
02

Find potential energy at the position

The formula for potential energy in SHM is \(U = \frac{1}{2} k x^2\), where \(k\) is the spring constant, and \(x\) is the displacement. Insert \(x = \frac{A}{2}\) in this formula to find potential energy.
03

Express the total energy in terms of the amplitude

We know that the total energy \(E\) in S.H.M. is given by \(E = \frac{1}{2} k A^2\), where \(A\) is the amplitude.
04

Isolate for the kinetic energy

Substitute the expressions for \(E\) and \(U\) in the kinetic energy formula (step1): \(K = E - U = \frac{1}{2} k A^2 - \frac{1}{2} k(\frac{A}{2})^2 = \frac{3}{4} * \frac{1}{2} k A^2\)
05

Substitute and simplify

Substitute \(\frac{1}{2} k A^2\) using \(E\), the total energy, in the expression found in Step 4. This gives \(K = \frac{3}{4} E\). Hence, when the displacement is half the amplitude, the kinetic energy is \(\frac{3}{4} E\).

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Most popular questions from this chapter

A particle is executing S.H.M. if its amplitude is \(2 \mathrm{~m}\) and periodic time 2 seconds. Then the maximum velocity of the particle will be (a) \(6 \pi\) (b) \(4 \pi\) (c) \(2 \pi\) (d) \(\pi\)

The bob of a simple pendulum of mass \(m\) and total energy \(E\) will have maximum linear momentum equal to [MP PMT 1986] (a) \(\sqrt{\frac{2 E}{m}}\) (b) \(\sqrt{2 m E}\) (c) \(2 m E\) (d) \(m E^{2}\)

When the potential energy of a particle executing simple harmonic motion is one-fourth of the maximum value during the oscillation, its displacement from the equilibrium position in terms of amplitude ' \(a\) ' is(a) \(a / 4\) (b) \(a / 3\) (c) \(a / 2\) (d) \(2 a / 3\)

A particle is executing S.H.M. if its amplitude is \(2 \mathrm{~m}\) and periodic time 2 seconds. Then the maximum velocity of the particle will be (a) \(6 \pi\) (b) \(4 \pi\) (c) \(2 \pi\) (d) \(\pi\)

A spring of force constant \(k\) is cut into two pieces such that one pieces is double the length of the other. Then the long piece will have a force constant of (a) \(2 / 3 k\) (b) \(3 \angle 2 k\) (c) \(3 k\) (d) \(6 k\)
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