Question

A simple pendulum consists of a ball of mass 4.6 kg hanging from a uniform string of mass 0.0601 g and length L. The period of oscillation for the pendulum is 1.43 s. Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Answer 1

Answer:

612.9 m/s

Explanation:

We are given that

Mass=M=4.6 kg

Mass of string=m=0.0601 g=$0.0601\times 10^{-3} kg$

$1 kg=10^3 g$

Length of pendulum=L

Time period=T=1.43 s

We have to determine the speed of transverse wave in the string.

Length of pendulum=$L=\frac{T^2g}{4\pi^2}$

Where $g=9.8m/s^2$

$\pi=3.14$

T=Time period

Length of pendulum=$L=\frac{(1.43)^2\times 9.8}{4\times (3.14)^2}$

Length of pendulum=L=0.51 m

Mass per unit length=$\mu=\frac{m}{L}=\frac{0.0601\times 10^{-3}}{0.51}=1.2\times 10^{-4}kg/m$

Tension in the string=$Mg=4.6\times 9.8=45.08 N$

Speed of the wave=$v=\sqrt{\frac{T}{\mu}}$

Where T=Tension in the string

Using the formula

Speed of the transverse wave=$v=\sqrt{\frac{45.08}{1.2\times 10^{-4}}}$m/s

Speed of the transverse wave=612.9 m/s