A composite string is made up by joining two | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The power transmitted by a wave is given by $P = \frac{1}{2} \mu \omega^2 A^2 v$,where $v = \sqrt{T/\mu}$.Let the lighter string have mass per uni

Vedclass · Accepted Answer

$89$

Vedclass · Answer

(B) The power transmitted by a wave is given by $P = \frac{1}{2} \mu \omega^2 A^2 v$,where $v = \sqrt{T/\mu}$.Let the lighter string have mass per unit length $\mu_1 = \mu$ and the heavier string have $\mu_2 = 4\mu$.The impedance of the strings are $Z_1 = \sqrt{\mu_1 T} = \sqrt{\mu T}$ and $Z_2 = \sqrt{\mu_2 T} = \sqrt{4\mu T} = 2\sqrt{\mu T}$.The amplitude of the transmitted wave $A_t$ is given by $A_t = \frac{2 Z_1}{Z_1 + Z_2} A_i = \frac{2 \sqrt{\mu T}}{\sqrt{\mu T} + 2\sqrt{\mu T}} A_i = \frac{2}{3} A_i$.The power transmitted $P_t$ is $P_t = \frac{1}{2} \mu_2 \omega^2 A_t^2 v_2$.Substituting $v_2 = \sqrt{T/4\mu} = \frac{1}{2} \sqrt{T/\mu} = \frac{1}{2} v_1$ and $A_t = \frac{2}{3} A_i$:$P_t = \frac{1}{2} (4\mu) \omega^2 (\frac{4}{9} A_i^2) (\frac{1}{2} v_1) = \frac{8}{9} (\frac{1}{2} \mu \omega^2 A_i^2 v_1) = \frac{8}{9} P_i$.The percentage of power transmitted is $\frac{8}{9} 	imes 100 \approx 88.89 \% \approx 89 \%$.

Similar Questions

$Assertion :$ The pitch of wind instruments rises and that of string instruments falls as an orchestra warms up.
$Reason :$ When temperature rises,the speed of sound increases,but the speed of a wave in a string fixed at both ends decreases.

$A$ wave pulse on a string has the dimensions shown in the figure. The wave speed is $v = 1 \, cm/s$. If point $O$ is a free end,what is the shape of the wave at time $t = 3 \, s$?

The superposing waves are represented by the following equations: ${y_1} = 5\sin 2\pi (10t - 0.1x)$ and ${y_2} = 10\sin 2\pi (20t - 0.2x)$. The ratio of intensities $\frac{I_{\max}}{I_{\min}}$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module