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Chapter 11: Problem 11

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) \(\pi\)

Short Answer

Expert verified
(c) \(2\pi\)

Step by step solution

01

Calculate the Angular Frequency

The angular frequency \(\omega\) is calculated using the formula \(\omega = \frac{2\pi}{T}\), where \(T\) is the given periodic time. For this case, \(T = 2\) seconds, hence \(\omega = \frac{2\pi}{2} = \pi\) rad/s.
02

Determine the Maximum Velocity

The maximum velocity \(v_{max}\) of the particle executing S.H.M. is calculated using the formula \(v_{max} = \omega \times a\), where \(a\) is the amplitude. For this case, \(a = 2\) m and \(\omega = \pi\) rad/s (from Step 1), hence \(v_{max} = \pi \times 2 = 2\pi\) m/s.
03

Identify the Answer

The maximum velocity value calculated in Step 2 can be seen as option (c) in the multiple-choice question. Therefore, the correct answer is (c) \(2\pi\).

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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 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) \(\sqrt{3} E / 4\)

A body is executing simple harmonic motion with an angular frequency \(2 \mathrm{rad} / \mathrm{sec}\). The velocity of the body at \(20 \mathrm{~mm}\) displacement. When the amplitude of motion is \(60 \mathrm{~mm}\) is [AFMC 1998] (a) \(40 \mathrm{~mm} / \mathrm{sec}\) (b) \(60 \mathrm{~mm} / \mathrm{sec}\) (c) \(113 \mathrm{~mm} / \mathrm{sec}\) (d) \(120 \mathrm{~mm} / \mathrm{sec}\)

A particle is executing simple harmonic motion with frequency \(f\). The frequency at which its kinetic energy changes into potential energy is (a) \(f / 2\) (b) \(f\) (c) \(2 f\) (d) \(4 f\)

The displacement of an oscillating particle varies with time (in seconds) according to the equation. \(y(c m)=\sin \frac{\pi}{2}\left(\frac{t}{2}+\frac{1}{3}\right)\). The maximum acceleration of the particle approximately (a) \(5.21 \mathrm{~cm} / \mathrm{sec}^{2}\) (b) \(3.62 \mathrm{~cm} / \mathrm{sec}^{2}\) (c) \(1.81 \mathrm{~cm} / \mathrm{sec}^{2}\) (d) \(0.62 \mathrm{~cm} / \mathrm{sec}^{2}\)
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