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

a=-((π2/64)) x ⇒ ω=√(π2/64)=(π/8) ⇒ T=(2 π/ω)=16 sec . There is a time difference of T / 2 between t=2 sec to t=10 sec . Hence particle is again passing through the mean position of SHM where its speed is maximum. i.e., 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

a=−(64π2​)x ⇒ω=64π2​​=8π​ ⇒T=ω2π​=16sec. There is a time difference of T/2 between t=2sec to t=10sec. Hence particle is again passing through the mean position of SHM where its speed is maximum. i.e., Vmax​=Aω=42​ ⇒A=π/842​​=π322​​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:

$\frac{32 2}{π} m$

$\frac{16}{π} m$

$\frac{4}{π} m$

$\frac{16 2}{π} m$