Simple Harmonic Motion Ques 66
- A particle executes simple harmonic motion between $x=-A$ and $x=+A$. The time taken for it to go from $O$ to $A / 2$ is $T _1$ and to go from $A / 2$ to $A$ is $T _2$, then
(2001, 2M)
(a) $T _1<T _2$
(b) $T _1>T _2$
(c) $T _1=T _2$
(d) $T _1=2 T _2$
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Answer:
Correct Answer: 66.(a)
Solution:
- In SHM, velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme positions. Therefore, the time taken for the particle to go from $O$ to $A / 2$ will be less than the time taken to go it from $A / 2$ to $A$, or $T _1<T _2$.
From the equations of SHM we can show that
$ t _1=T _{0-A / 2}=T / 12 \text { and } t _2=T _{A / 2-A}=T / 6 $
$ \text { So, that } t _1+t _2=T _{O-A}=T / 4 $
BETA
