A particle performs simple harmonic motion with amplitude A- Its speed is trebled at the instant that it is at a distance 2 A -3 from equilibrium position- The new amplitude of the motion is-A- 3 AB- A -3C- 7 -D- A-3-41

The correct option is C v=∞√(A^2 (2A/3)^2)= ∞√(5)A/3 Now, V'=3 ×∞√(5)A/3 Now, V'= ∞√(A'^2 (2A/3)^2)⇒ A' = 7A/3 ...

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Standard XII

Physics

SHM Formulae

A particle pe...

Question

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is:

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Solution

The correct option is C

$v = \infty \sqrt{A^{2} - {(\frac{2 A}{3})}^{2}} = \frac{\infty \sqrt{5} A}{3} N o w, V^{'} = 3 \times \frac{\infty \sqrt{5} A}{3} N o w, V^{'} = \infty \sqrt{A^{' 2} - {(\frac{2 A}{3})}^{2}} \Rightarrow A^{'} = \frac{7 A}{3}$

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A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A3 from equilibrium position. The new amplitude of the motion is:

The correct option is C v=∞√A2−(2A3)2=∞√5A3Now, V′=3×∞√5A3Now, V′=∞√A′2−(2A3)2⇒A′=7A3

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is:

The correct option is C

$v = \infty \sqrt{A^{2} - {(\frac{2 A}{3})}^{2}} = \frac{\infty \sqrt{5} A}{3} N o w, V^{'} = 3 \times \frac{\infty \sqrt{5} A}{3} N o w, V^{'} = \infty \sqrt{A^{' 2} - {(\frac{2 A}{3})}^{2}} \Rightarrow A^{'} = \frac{7 A}{3}$