A particle is executing SHM of period $24x\ $and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q. where OP = $- 9$ cm and OQ =40 cm is

Question

Dr. Abubacker · Accepted Answer

(b) For displacement $OQ = 40$ cm; let $t_{1}$ be the time taken then
$40 = 41\sin\frac{2\pi}{24}t_{1},$ on solving $t_{1} = 5.16$ s
For displacement $OQ = - 9$ cm, let $t_{2}$ be the time taken then $9 = 41\sin\frac{2\pi}{12}t_{2}$,
On solving $t_{2} = 0.84$ s
Total time $= 5.16 + 0.84 = 6.00$ s

Q. A particle is executing SHM of period $24x\ $and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q. where OP = $- 9$ cm and OQ =40 cm is

Explanation

Reviewed by Dr. Abubacker

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