MHT CET · Physics · Oscillations
A particle performing linear S.H.M. has period 8 second. At time \(t=0\), it is in the mean position. The ratio of the distances travelled by the particle in the \(1^{\text {st }}\) and \(2^{\text {nd }}\) second is \(\left(\cos 45^{\circ}=1 / \sqrt{2}\right)\)
- A \(1:(\sqrt{2}-1)\)
- B \(1: 2\)
- C \(2: 1\)
- D \(1:(\sqrt{2}+1)\)
Answer & Solution
Correct Answer
(A) \(1:(\sqrt{2}-1)\)
Step-by-step Solution
Detailed explanation
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4}\) \(x(t) = A \sin(\omega t)\)
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