Question

consider a situation of simple harmonic motion in which the distance between the endpoints is 2.99 m and exactly 7 cycles are completed in 21.9 s. when this motion is viewed as a projection of circular motion, what is the radius, r, and angular velocity, ω, of the circular motion?
r =

m
ω =

rad/s

Explanation:

Step1: Find the radius

In simple harmonic motion (viewed as a projection of circular motion), the distance between the endpoints is \(2r\). Given the distance between endpoints \(d = 2.99\space m\).

$$r=\frac{d}{2}$$

$$r=\frac{2.99}{2}=1.495\space m$$

Step2: Find the angular velocity

The period \(T\) of the simple - harmonic motion (which is the same as the period of the circular motion) is given by \(T=\frac{t}{N}\), where \(t = 21.9\space s\) is the total time and \(N = 7\) is the number of cycles.

$$T=\frac{21.9}{7}=3.1286\space s$$

The formula for angular velocity \(\omega\) is \(\omega=\frac{2\pi}{T}\)

$$ \omega=\frac{2\pi}{3.1286}$$

$$ \omega\approx2.01\space rad/s$$

Answer:

\(r = 1.50\space m\) (rounded to three significant figures), \(\omega\approx2.01\space rad/s\)