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=$\frac{372}{1000}=0.372 Kg$

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=$\omega A$

$17.6=A\omega^2$

$1.75=A\omega$

$\frac{17.6}{1.75}=\frac{A\omega^2}{A\omega}=\omega$

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

b.Substitute the value of angular frequency

$1.75=A(10.06)$

$A=\frac{1.75}{10.06}=0.17 m$

Hence, the amplitude=0.17 m

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

Using the formula

$k=0.372\times (10.06)^2$

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