A1.2 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

(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

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

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

Put the value into the formula

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

Put the value into the equation

The expression for x as a function of time is

Hence, This is the required solution.

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distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000