Question

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 1.9 m/s. Determine the amplitude A of the motion. A = Entry field with correct answer .07756

Answer 1

The amplitude of the motion is 0.077 m

Step-by-step explanation:

It is given that,

Mass of the object, m = 0.2 kg

Spring constant, K = 120 N/m

Maximum speed of the object, v = 1.9 m/s

We know that, in harmonic motion, the maximum speed of the object is given by :

`v_(max)=A\omega`

Where

A is the amplitude of the wave

`\omega` is the angular velocity,
`\omega=\sqrt{(k)/(m)}`

`v_(max)=A\sqrt{(k)/(m)}`

`A=v_(max)\sqrt{(m)/(K)}`

`A=1.9* \sqrt{(0.2)/(120)}`

A = 0.077 m

So, the amplitude of the motion is 0.077 m. Hence, this is the required solution.

Answer 2

0.07756 m

Step-by-step explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by
`v_(max)=A\omega` where A is amplitude and
`\omega` is angular velocity

Angular velocity is given by
`\omega=\sqrt{(k)/(m)}` where k is spring constant and m is mass

So
`v_(max)=A\sqrt{(k)/(m)}`

`A=V_(max)\sqrt{(m)/(k)}=1.9* \sqrt{(0.2)/(120)}=0.07756 \ m`