Question

1. Choose the correct option.
i) A particle performs linear S.H.M.
starting from the mean position. Its
amplitude is A and time period is T. At
the instance when its speed is half the
maximum speed, its displacement x is
13
А A
(A) А A
2
(c) 42 (D) A
2
2.
sl-will
1
i)
ii
​

Answer 1

Answer:

The relation between angular frequency and displacement is given as

$v = w \sqrt{ {a}^{2} - {x}^{2} }$

Suppose

$x = aw \sin\: wt$

On differentiating the above equation w.r.t. time we get

$\frac{dr}{dt} a \: w \: cos \: wt$

The maximum value of velocity will be\[{v_{\max }} = A\omega \]

The displacement for the time when speed is half the maximum is given as :-

$v = \frac{aw}{2}$

${a}^{2} {w}^{2} 4w( {a}^{2} - {x}^{2} )$

By substituting the value in(1)we get the displacement as:-

$x = \frac{a \sqrt{3} }{2}$