Question

0.96 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 310 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.9 s is closest to: 0.64 m/s

-0.90 m/s
0.90 m/s
-0.64 m/s
zero

Answer 1

The velocity of the block is 0.64 m/s.

(A) is correct option.

Step-by-step explanation:

Given that,

Mass of block = 0.96 kg

Force constant = 310 N/m

Amplitude = 0.080 m

Time = 0.9

We need to calculate the velocity of the block

We know that,

Equation of motion

`x(t)=A\cos(\omega t)`

On differentiating

`(d(x(t)))/(dt)=-0.080\omega \sin(\omega t)`

`v(t)=-0.080\omega \sin(\omega t)`....(I)

We need to calculate the
`\omega`

`\omega=\sqrt{(k)/(m)}`

Put the value into the formula

`\omega=\sqrt{(310)/(0.96)}`

`\omega=17.96\ rad/s`

Put the value of
`\omega` in equation (I)

`v(t)=-0.080*17.97 \sin(17.97*0.9)`

`v(t)=0.64\ m/s`

Hence, The velocity of the block is 0.64 m/s.