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

Answer:

465.6 N/m

Explanation:

We are given that

F=12 N

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

We have to find the spring constant of the spring.

$F=mg$

Where $g=9.8 m/s^2$

Using the formula

$12=m\times 9.8$

$m=\frac{12}{9.8}$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,$k=\frac{12}{9.8}\times (19.5)^2=465.6 N/m$