A particle executes simple harmonic motion between x = – A and x= +A. The time taken for it to go from 0 to A2 is T1 and to go from A2 to A is T2. Then:

# A particle executes simple harmonic motion between x = – A and x= +A. The time taken for it to go from 0 to $\frac{\mathrm{A}}{2}$ is ${\mathrm{T}}_{1}$ and to go from $\frac{\mathrm{A}}{2}$ to A is ${\mathrm{T}}_{2}$. Then:

1. A

${\mathrm{T}}_{1}<{\mathrm{T}}_{2}$

2. B

${\mathrm{T}}_{1}>{\mathrm{T}}_{2}$

3. C

${\mathrm{T}}_{1}={\mathrm{T}}_{2}$

4. D

${\mathrm{T}}_{1}=2{\mathrm{T}}_{2}$

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### Solution:

In simple harmonic motion, the speed of the particle is the maximum at the mean position x = 0, decreases as it moves towards the extreme position becoming zero at the extreme position x = A. Hence the  particle will take shorter time to move from x = 0 to x = $\frac{\mathrm{A}}{2}$ than to move from x = $\frac{\mathrm{A}}{2}$ to x = A.  Register to Get Free Mock Test and Study Material

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