First slide

Simple hormonic motion

Question

A particle executing S.H.M. between x = - A and x = + A. {. The time taken for it to go from 0 to A /2 is T₁and to go from A /2 to A is T₂.Then

Easy

A

T₁<T₂
B

T₁>T₂
C

T₁ =T₂
D

T₁=2T₂

Solution

We know that x=A sin $ωt$ i.e., x=0 at t =0
$Now \frac{A}{2} = Asin {ωT}_{1} or T_{1} = π / 6 ω Furthor T_{2} = \frac{T}{4} - T_{1} = \frac{2 π}{4 ω} - \frac{π}{6 ω} = \frac{2 π}{6 ω} = \frac{π}{3 ω} T_{1} < T_{2}$

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