Question

Following graph shows a particle performing S.H.M. about mean position \( x=0 \). The equation of particle if \( t =\frac{ T }{4} \) is taken as starting time is (Notations have usual meanings)

(A) \( A \sin \left(\omega t+\frac{2 \pi}{3}\right) \)

(B) \( A \sin \left(\omega t+\frac{\pi}{3}\right) \)

(C) \( A \sin \left(\omega t +\frac{\pi}{6}\right) \)

(D) \( A \cos \left(\omega t+\frac{2 \pi}{3}\right) \)

Answer 1

Correct option is (B) \(A \sin (\omega t + \frac \pi 3)\)

Let \(x = A \sin(\omega t + \phi)\)

Given at \(t = \frac T4, x = \frac A2\)

\(\therefore \frac A2 = A \sin(\omega\frac T4 + \phi)\)

\(\Rightarrow \frac 12 = \sin (\frac {2\pi}4 + \phi)\)

\(\Rightarrow \frac 12 = \sin (\frac {\pi}2 + \phi)\)

\(\Rightarrow \frac 12= \cos\phi\)

\(\Rightarrow \phi = \frac \pi 3\)

\(\therefore x = A \sin (\omega t + \frac \pi 3)\)