Question

In the previous question, the percentage of power transmitted to the heavier string through the joint is approximately

• A. 33%
• B. 89%
• C. 67%
• D. 75%

Answer 1

The correct option is B 89%

Power, $P=\frac{1}{2}\mu {\omega }^2{A}^{2}v$
Amplitude of transmitted wave, $A_t=\frac{2Z_1}{Z_1+Z_2}A=\frac{2\sqrt{\mu T}}{\sqrt{\mu T}+\sqrt{4\mu T}}A=\frac{2}{3}A$
Thus, $P=\frac{1}{2}\mu {\omega }^2{A}^{2}v=\frac{1}{2}\times 4\mu \times {\omega }^2\times \frac{4A^2}{9}\times \sqrt{\frac{T}{4\mu }}=(\frac{1}{2}\times \mu \times {\omega }^2\times A^2\times \sqrt{\frac{T}{\mu }})\times \frac{8}{9}=\frac{8}{9}\times InitialpowerThuspowertransmitted=\frac{8}{9}\times 100=89$%