A string of mass 3 g and length 1 m can carry a wave of 20 mW power.When a tension of 15 N is applied to the string it oscillates with a period of 40 ms.The amplitude of the wave is
A)2.76 mm.
B)9.80 mm.
C)11.8 mm.
D)18.5 mm.
E)23.5 mm.

Question

Quizplus · Accepted Answer

The amplitude of the wave can be calculated using the formula for the power of a wave on a string, which is given by $P = \frac{1}{2} \mu A^2 \omega^2 v$, where $P$ is the power of the wave, $\mu$ is the mass per unit length of the string, $A$ is the amplitude of the wave, $\omega$ is the angular frequency, and $v$ is the velocity of the wave on the string.Given:- $P = 20 \, \text{mW} = 20 \times 10^{-3} \, \text{W}$- Mass of the string, $m = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg}$- Length of the string, $L = 1 \, \text{m}$- Tension in the string, $T = 15 \, \text{N}$- Period of oscillation, $T_{\text{period}} = 40 \, \text{ms} = 40 \times 10^{-3} \, \text{s}$First, calculate the mass per unit length ($\mu$) of the string:$\mu = \frac{m}{L} = \frac{3 \times 10^{-3} \, \text{kg}}{1 \, \text{m}} = 3 \times 10^{-3} \, \text{kg/m}$The velocity ($v$) of the wave on the string can be found using the formula $v = \sqrt{\frac{T}{\mu}}$:$v = \sqrt{\frac{15}{3 \times 10^{-3}}} = \sqrt{5000} \, \text{m/s}$The angular frequency ($\omega$) can be calculated from the period:$\omega = \frac{2\pi}{T_{\text{period}}} = \frac{2\pi}{40 \times 10^{-3}} = 50\pi \, \text{rad/s}$Substituting these values into the power formula and solving for $A$:$20 \times 10^{-3} = \frac{1}{2} \times 3 \times 10^{-3} \times A^2 \times (50\pi)^2 \times \sqrt{5000}$Solving for $A$, we find:$A \approx 2.76 \, \text{mm}$Therefore, the correct answer is A) 2.76 mm.

A String of Mass 3 G and Length 1 M