Question

A 4 kg particle is moving along the x-axis under the action of the force F = -$\left( \frac { \pi ^ { 2 } } { 16 } \right)$x N. When t = 2s, the particle passes through the origin and when t = 10s, its speed is 4√2 m/s. The amplitude of the motion is :

• A. $\frac { 32 \sqrt { 2 } } { \pi }$m/s
• B. $\frac { 16 } { \pi }$m/s
• C. $\frac { 4 } { \pi }$m/s
• D. $\frac { 16 \sqrt { 2 } } { \pi }$m/s

Answer 1

Answer: (A)

$\frac { 32 \sqrt { 2 } } { \pi }$m/s

Answer 2

a = - $\left( \frac { \pi ^ { 2 } } { 64 } \right)$x

⇒ ω = $\sqrt { \frac { \pi ^ { 2 } } { 64 } } = \frac { \pi } { 8 }$

T = $\frac { 2 \pi } { \omega }$ =16 sec.

There is a time difference of between t = 2 sec. to

t = 10 sec. Hence particle is again passing through mean

position of SHM where its speed is maximum.

i.e. V_max = Aω = 4√2

A = $\frac { 32 \sqrt { 2 } } { \pi }$m.