IIT-JEE 2009 · Paper 1 · Physics Q48

PhysicsSingle correct+3 / -1Question 48 of 60

The $x$ - $t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t=4/3$ s is

\dfrac{\sqrt3}{32}\pi^2\ \mathrm{cm/s^2}

\dfrac{-\pi^2}{32}\ \mathrm{cm/s^2}

\dfrac{\pi^2}{32}\ \mathrm{cm/s^2}

-\dfrac{\sqrt3}{32}\pi^2\ \mathrm{cm/s^2}

Answer and worked solution

Official answer: D

From the graph. Amplitude 1 cm, period 8 s, starting at $x=0$ moving up: $x=\sin\dfrac{\pi t}4$ cm, $\omega=\dfrac\pi4$ .

Acceleration. $a=-\omega^2x=-\dfrac{\pi^2}{16}\sin\dfrac\pi3=-\dfrac{\sqrt3}{32}\pi^2\ \mathrm{cm/s^2}$ at $t=\frac43$ s.

(D)

Topics:Simple harmonic motion

IIT-JEE 2009 · Paper 1, Physics question 48