A 4 kg particle is moving along the x -axis under the action of the force F=-((π2/16)) x N . At t=2 sec, the particle passes through the origin and at t=10 sec its speed is 4 √2 m / s. The amplitude of the motion is:

Question

A 4 kg particle is moving along the x -axis under the action of the force F=-((π2/16)) x N . At t=2 sec, the particle passes through the origin and at t=10 sec its speed is 4 √2 m / s. The amplitude of the motion is: (A) (32 √2/π) m (B) (16/π) m (C) (4/π) m (D) (16 √2/π) m

Tardigrade Admin · Accepted Answer

Acceleration of particle is a=-((π2/64)) x comparing with a=-ω2 x we get ⇒ ω=√(π2/64)=(π/8) ⇒ T=(2 π/ω)=16 s There is a time difference of (T/2) between t=2 s to t=10 s. Hence particle is again passing through the mean position of SHM where its speed is maximum at t=10 s ⇒ V max =A ω=4 √2 ⇒ A=(4 √2/π / 8) =(32 √2/π) m .

Q. A $4 k g$ particle is moving along the $x$ -axis under the action of the force $F = - (\frac{π ^{2}}{16}) x N .$ At $t = 2 sec$ , the particle passes through the origin and at $t = 10 sec$ its speed is $42 m / s$ . The amplitude of the motion is:

$\frac{32 2}{π} m$

$\frac{16}{π} m$

$\frac{4}{π} m$

$\frac{16 2}{π} m$

Q. A 4kg particle is moving along the x -axis under the action of the force F=−(16π2​)xN. At t=2sec, the particle passes through the origin and at t=10sec its speed is 42​m/s. The amplitude of the motion is:

π322​​m

π16​m

π4​m

π162​​m

Q. A $4 k g$ particle is moving along the $x$ -axis under the action of the force $F = - (\frac{π ^{2}}{16}) x N .$ At $t = 2 sec$ , the particle passes through the origin and at $t = 10 sec$ its speed is $42 m / s$ . The amplitude of the motion is:

$\frac{32 2}{π} m$

$\frac{16}{π} m$

$\frac{4}{π} m$

$\frac{16 2}{π} m$