JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Elasticity & SHM

The

x

t

graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at

t = 4/3 s

Answer: (D)

Solution

From the graph, the motion is

x = A sin (ω t)

where

A = 1

cm and the period

T = 8

s, so

ω = \frac{2 π}{T} = \frac{π}{4}

rad/s. The displacement is

x = sin (\frac{π}{4} t)

. The acceleration is

a = - ω^{2} x = - (\frac{π}{4})^{2} sin (\frac{π}{4} t) = - \frac{π ^{2}}{16} sin (\frac{π}{4} t)

. At

t = 4/3

sin (\frac{π}{4} \cdot \frac{4}{3}) = sin (\frac{π}{3}) = \frac{3}{2}

. Therefore,

a = - \frac{π ^{2}}{16} \cdot \frac{3}{2} = - \frac{3 π ^{2}}{32}

cm/s².

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option