First slide

Simple hormonic motion

Question

A particle executes S.H.M. of amplitude A. If I₁, and T₂, are the times taken by the particle to traverse from 0 to A/2 and from A I 2 to A respectively, then(T₁ / T₂) will be equal to

Easy

A

1
B

1 /2
C

1 /4
D

2

Solution

For S.H.M.
$x = Asin ωt = Asin (2 πt / T) he time taken by the particle for travelling from x = 0 to x x = Al 2 isTt . Hence \frac{A}{2} = Asin (\frac{2 {πT}_{1}}{T}) or \sin (\frac{2 {πT}_{1}}{T}) = \frac{1}{2} or (\frac{2 {πT}_{1}}{T}) = \frac{π}{2} or T_{1} = (\frac{T}{12}) Similarly, time taken bythe particle to traverse from x = 0 to x =, 4 i . e ., one - fourth cycle is \frac{T}{4} = T_{1} + T_{2} T_{2} = \frac{T}{4} - T_{1} = \frac{T}{4} - \frac{T}{12} = (\frac{T}{6}) From eqs. (1) and (2), we get \frac{T_{1}}{T_{2}} = \frac{(T / 12)}{T / 6} = \frac{1}{2}$

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