Question

Two instruments having stretched strings are being played in unison. When the tension of one of the instruments is increased by $1$%, $3$ beats are produced in $2s$. The initial frequency of vibration of each wire is

• A. $300Hz$
• B. $500Hz$
• C. $1000Hz$
• D. $400Hz$

Answer 1

Answer: (A)

$300Hz$

Beats= difference in frequencies.
Frequency $n=\frac{1}{2l}\sqrt{\frac{T}{m}}$ where T is the tension and m is the mass.
Tension of one of the instruments is increased by 1%.
3 beats are produced in 2 seconds.
Therefore new Tension $T_1=T+0.01T=1.01T$
Difference in frequencies = $n_1-n=\frac{3}{2}$
$\frac{1}{2l}\sqrt{\frac{1.01T}{m}}-\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{3}{2}$
$\therefore$ $1.005n-n=\frac{3}{2}$
n=300 Hz.