Question

A particle executes S.H.M. between \(x=-A\) and \(x=+A\). The time taken for it to go from mean position O to \(\frac{A}{2}\) is \(T_{1}\) and to go from \(\frac{A}{2}\) to \(A\) is \(T_{2}\), then (\(\sin 30^{\circ }=1/2\), \(\sin 90^{\circ }=1\))

• A. \(T_{1}=2T_{2}\)
• B. \(T_{1}=T_{2}\)
• C. \(T_{1}>T_{2}\)
• D. \(T_{1}\lt T_{2}\)

Hint:

Use x equal to A sin omega t to find the times, then compare the two intervals.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Starting from the mean position, \(x=A\sin\omega t\).

Step 2: Time to reach A/2
\(\dfrac{A}{2}=A\sin\omega T_{1}\), so \(\sin\omega T_{1}=\dfrac{1}{2}\) and \(\omega T_{1}=30^{\circ}=\dfrac{\pi}{6}\). Hence \(T_{1}=\dfrac{T}{12}\).

Step 3: Time to reach A
\(A\) is reached at \(\omega t=90^{\circ}\), i.e. \(t=\dfrac{T}{4}\). So \(T_{2}=\dfrac{T}{4}-\dfrac{T}{12}=\dfrac{T}{6}\).
Hence \(T_{1}<T_{2}\).

Final Answer:
T one is less than T two. \[ \boxed{\text{(D) }T_{1}\lt T_{2}} \]