For Original SHM,
using v=ωA2−X2
v=ωA2−94A2
=3ω5A
New SHM will be,
3v=ωAN2−XN2
33ω5A=ωAN2−94A2
5A2=AN2−94A2
AN2=949A2
AN=37A
A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 32A from equilibrium position. The new amplitude of the motion is:
Held on 3 Apr 2016 · Verified 6 Jul 2026.
A3
37A
3A41
3A
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Match List-I with List-II. <table class="pyq-table"><tbody><tr><th>List-I</th><th>List-II</th></tr><tr><td>A. $\sin^2 \omega t$</td><td>I. Periodic with time period $T = \dfrac{\pi}{\omega}$ but not simple harmonic motion (SHM)</td></tr><tr><td>B. $\sin^3(2\omega t)$</td><td>II. Periodic with time period $T = \dfrac{2\pi}{\omega}$ but Not SHM</td></tr><tr><td>C. $\sin(\omega t) + \cos(\pi \omega t)$</td><td>III. Periodic with time period $T = \dfrac{\pi}{\omega}$ and SHM</td></tr><tr><td>D. $\cos\omega t + \cos 2\omega t$</td><td>IV. Non-periodic</td></tr></tbody></table> Choose the correct answer from the options given below :
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