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

The frequency of the block's motion is 10 Hz and the amplitude is approximately 0.0827 m. The expression for x as a function of time is x(t) = 0.0827 * cos(20πt).

Step-by-step explanation:

The frequency and amplitude of the block's motion can be determined using the equations of motion for a block and spring system. The frequency, denoted as f, is given by the equation:

f = 1 / (2π) * sqrt(k / m)

where k is the force constant of the spring (480 N/m) and m is the mass of the block (1.2 kg). Plugging in the values, we get:

f = 1 / (2π) * sqrt(480 / 1.2) = 10 Hz

The amplitude, denoted as A, is the maximum displacement of the block from its equilibrium position. In this case, the block starts at x = 0 with a speed of 5.2 m/s. This gives us the initial velocity, which we can use to find the amplitude:

v = ωA
A = v / ω
where ω is the angular frequency, given by ω = 2πf. Plugging in the values, we get:

ω = 2π * 10 = 20π
A = 5.2 / (20π) ≈ 0.0827 m

So, the amplitude of the block's motion is approximately 0.0827 m.

The expression for x as a function of time, denoted as x(t), can be obtained using the equation of motion for simple harmonic motion:

x(t) = A * cos(ωt)

where A is the amplitude and ω is the angular frequency. Plugging in the values, we get:

x(t) = 0.0827 * cos(20πt)

Answer 2

(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+(\pi)/(2))`

Step-by-step 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=(1)/(2\pi)\sqrt{(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

`(1)/(2)* kA^2=(1)/(2)* mv^2`

Put the value into the formula

`(1)/(2)*500* A^2=(1)/(2)*1.2*(5.2)^2`

`A^2=(1.2*(5.2)^2)/(500)`

`A=\sqrt{(1.2*(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+(\pi)/(2))`

The expression for x as a function of time is
`x=0.255\cos(19.9 t+(\pi)/(2))`

Hence, This is the required solution.