Oscillations · Question 5

A block of mass 5 kg moves along the x-direction subject to the force F=(-20x+10) N, with the value of x in metre. At time t=0 s, it is at rest at position x=1 m. The position and momentum of the block at t=( /4) s are

1. -0.5 m, 5 kg m/s
2. 0.5 m, 0 kg m/s
3. 0.5 m, -5 kg m/s
4. -1 m, 5 kg m/s

Answer

(3) 0.5 m, -5 kg m/s

Solution

F=-20(x-0.5) ⇒ SHM about x0=0.5 m with = 20/5=2 rad s-1.

Starts from rest at x=1 m ⇒ amplitude A=0.5 m: x=0.5+0.5 2t.

At t= /4: ( /2)=0 ⇒ x=0.5 m (mean position).

v=-0.5 2 2t=-1 m s-1 ⇒ p=5 (-1)=-5 kg m s-1.

Answer: (C)