Question

A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by y(t)=4.50cmcos[(19.5s−1)t−π/8].(a) What is the spring constant of the spring?

Answer 1

465.6 N/m

Step-by-step explanation:

We are given that

F=12 N

`y(t)=4.50cos\left \{(19.5s^(-1)t-(\pi)/(8)\right \}`

We have to find the spring constant of the spring.

`F=mg`

Where
`g=9.8 m/s^2`

Using the formula

`12=m* 9.8`

`m=(12)/(9.8)`kg

Compare the given equation with

`y(t)=Acos(\omega t-\phi)`

We get
`\omega=19.5`

`k=m\omega^2`

Using the formula

Spring constant,
`k=(12)/(9.8)* (19.5)^2=465.6 N/m`