First slide

Simple harmonic motion

Question

A particle execute SHM from the mean position. Its amplitude is A, its time period is ‘T’. At what displacement, its speed is half of its maximum speed.

Moderate

A

$\frac{\sqrt{3} A}{2}$
B

$\frac{\sqrt{2}}{3} A$
C

$\frac{2 A}{\sqrt{3}}$
D

$\frac{3A}{\sqrt{2}}$

Solution

$given oscillator speed is V = \frac{1}{2} V_{max} = \frac{1}{2} {Aω.........(1) where maximum velocity V}_{max} = A ω o s c i l l a t o r v e l o c i t y i s V = ω \sqrt{A^{2} {-Y}^{2}} ---(2) s u b s t i t u t e (1) i n (2) \frac{1}{2} Aω=ω \sqrt{A^{2} {-Y}^{2}} \Rightarrow \frac{1}{2} A= \sqrt{A^{2} {-Y}^{2}} \Rightarrow \frac{1}{4} A^{2} {=A}^{2} {-Y}^{2} \Rightarrow Y^{2} = A^{2} - \frac{A^{2}}{4} \Rightarrow Y^{2} = \frac{{3A}^{2}}{4} \Rightarrow Y = \frac{\sqrt{3} A}{2}$

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