Question

A 1.2 kg block sliding on a horizontal frictionless surface is attached to a horizontal spring with k =480 N/m. Let x be the displacement of the block from the position at which the spring is upstretched. At t = 0 the block passes through x = 0 with a speed of 5.2 m/s in the positive xdirection. What are the (a) frequency and (b) amplitude of the block’s motion? (c) Write an expression for x as a function of time

Answer 1

Answer:

(a). The frequency is 3.18 Hz.

(b). The amplitude of the block's motion is 0.255 m.

(c). The expression for x as a function of time is $x=0.255\cos(19.9 t+\dfrac{\pi}{2})$

Explanation:

Given that,

Mass of block = 1.2 kg

Spring constant = 480 N/m

Speed = 5.2 m/s

We need to calculate the frequency

Using formula of frequency

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{480}{1.2}}$

$f=3.18\ Hz$

The frequency is 3.18 Hz.

(b). We need to calculate the amplitude of the block's motion

Using relation of equation of amplitude and kinetic energy

$\dfrac{1}{2}\times kA^2=\dfrac{1}{2}\times mv^2$

Put the value into the formula

$\dfrac{1}{2}\times500\times A^2=\dfrac{1}{2}\times1.2\times(5.2)^2$

$A^2=\dfrac{1.2\times(5.2)^2}{500}$

$A=\sqrt{\dfrac{1.2\times(5.2)^2}{500}}$

$A=0.255\ m$

The amplitude of the block's motion is 0.255 m.

(c). We need to write the expression for x as a function of time

$x=A\cos(\omega t+\phi)$

Put the value into the equation

$x=0.255\cos(19.9 t+\dfrac{\pi}{2})$

The expression for x as a function of time is $x=0.255\cos(19.9 t+\dfrac{\pi}{2})$

Hence, This is the required solution.