Question

Two identical stringed instruments have frequency 100Hz. If tension in one of them is increased by 4% and they are sounded together, then number of beats in one second is :

• A. 1
• B. 3
• C. 4
• D. 2

Answer 1

Answer: (D)

2

Given : $\frac{T_1}{T_2}=\frac{100}{104}$ and $\delta =100H_z$

$\frac{{\delta }_1}{{\delta }_2}=\sqrt{\frac{T_1}{T_2}}$ $(\therefore \delta \propto \sqrt{T})$

$\frac{{\delta }_1}{{\delta }_2}=\sqrt{\frac{100}{104}}$

$\frac{{\delta }_1}{{\delta }_2}=0.98$

$\delta =\frac{{\delta }_1}{0.98}=\frac{100}{0.98}=102H_z$

$Beats={\delta }_2-{\delta }_1$ $=102-100$ $=2$