Each of the two strings of lengths $51.6 ext{ cm}$ and $49.1 ext{ cm}$ is tensioned separately by $20 ex

Question

Each of the two strings of lengths $51.6 	ext{ cm}$ and $49.1 	ext{ cm}$ is tensioned separately by $20 	ext{ N}$ of force. The mass per unit length of both strings is the same and equals $1 	ext{ g/m}$. When both the strings vibrate simultaneously, the number of beats is:

Accepted Answer

Given:
Lengths of the strings, $l_1 = 51.6 	ext{ cm} = 0.516 	ext{ m}$ and $l_2 = 49.1 	ext{ cm} = 0.491 	ext{ m}$
Tension in the strings, $T = 20 	ext{ N}$
Mass per unit length, $\mu = 1 	ext{ g/m} = 1 	imes 10^{-3} 	ext{ kg/m}$

The velocity of a transverse wave in a stretched string is given by:
$$v = \sqrt{\frac{T}{\mu}}$$
$$v = \sqrt{\frac{20}{1 	imes 10^{-3}}} = \sqrt{20000} \approx 141.42 	ext{ m/s}$$

The fundamental frequency of a vibrating string of length $l$ is $f = \frac{v}{2l}$.
The frequencies of the two strings are:
$$f_1 = \frac{v}{2l_1} = \frac{141.42}{2 	imes 0.516} = \frac{141.42}{1.032} \approx 137.03 	ext{ Hz}$$
$$f_2 = \frac{v}{2l_2} = \frac{141.42}{2 	imes 0.491} = \frac{141.42}{0.982} \approx 144.01 	ext{ Hz}$$

The number of beats produced per second is the difference between the two frequencies:
$$	ext{Number of beats} = f_2 - f_1 = 144.01 - 137.03 = 6.98 \approx 7$$

Step-by-Step Solution

Practice All PHYSICS Questions