Question

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answer 1

Answer with Explanation:

We are given that

Mass , m=372 g=
`(372)/(1000)=0.372 Kg`

1 kg=1000g

Maximum acceleration, a=
`17.6 m/s^2`

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a=
`A\omega^2`

Maximum speed, v=
`\omega A`

`17.6=A\omega^2`

`1.75=A\omega`

`(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega`

Angular frequency,
`\omega=10.06 rad/s`

b.Substitute the value of angular frequency

`1.75=A(10.06)`

`A=(1.75)/(10.06)=0.17 m`

Hence, the amplitude=0.17 m

c.Spring constant,k=
`m\omega^2`

Using the formula

`k=0.372* (10.06)^2`

Hence, the spring constant,k=37.6 N/m