First slide

Simple harmonic motion

Question

The velocity of particle undergoing SHM at the mean position is 4m/s. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude

Moderate

A

$2 \sqrt{3} m/s$
B

$5 \sqrt{3} m/s$
C

$4 \sqrt{3} m/s$
D

$3 \sqrt{3} m/s$

Solution

${maximum velocity of simple harmonic oscillator is V}_{max} =w \times A w h e r e w = a n g u l a r v e l o c i t y, A = a m p l i t u d e, g i v e n, wA=4m/s w e k n o w t h a t, V e l o c i t y o f o s c i l l a t o r i s V=w \sqrt{A^{2} {-x}^{2}} here x= displacement of the oscillator \Rightarrow V =w \sqrt{A^{2} {-A}^{2} / 4} \Rightarrow V =w \sqrt{3 A^{2} / 4} \Rightarrow V =w \sqrt{3} \frac{A}{2}, s u b s t i t u t e w A = 4 v = 2 \sqrt{3} m / s$

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