Question

A particle executes simple harmonic motion about the point X =0 . At time t =0 it has displacement cm and zero velocity. If the frequency of motion is $0.25se{c}^{-2}$, find
(a) the time period,
(b) angular frequency,
(c) the amplitude,
(d) maximum speed,
(e) the displacement at t =3 sec and
(f) the velocity at t =3 sec.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C

3

Time period $T=\frac{1}{n}=\frac{1}{0.25se{c}^{-1}}=4sec.$
(b) Angular frequency $\omega =\frac{2\pi }{T}=\frac{2\pi }{4}=\frac{\pi }{2}rad/sec=1.57rad/s.$
(c) Amplitude is the maximum displacement from mean position. Hence, A=2 -0 =2 cm
(d) Maximum speed $V_{max}=A\omega =2,\frac{\pi }{2}=\pi =\pi cm/s=3.14cm/s$
(e) The displacement is given by $x=Asin(ex+\varphi )$
At t =0, x =2cm, $\therefore 2=2sin\varphi$
or $sin\varphi =1=sin{90}^{\circ }or\varphi -{90}^{\circ }$
Now, at t=3 sec $\Rightarrow x=2sin(\frac{\pi }{2}\times 3+\frac{\pi }{2})=0$
(f) Velocity at x =0 is $V_{max}$, i.e. 3.14 cm/se