Question

Two identical wires are stretched by the same tension of 100 N and each emits a note of frequency 200 Hz. If the tension in one wire is increased by 1 N, then the beat frequency is

• A. $2Hz$
• B. $\frac{1}{2}Hz$
• C. $1Hz$
• D. $None$

Answer 1

Answer: (C)

$1Hz$

$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{f_1}{200}=\sqrt{\frac{101}{100}}$

$\frac{f_1}{200}=\sqrt{1.01}$

using binomial expansion

$\frac{f_1}{200}=1+\frac{.01}{2}$

$f_1=200+1Hz=201Hz$

beats frequency heard is $f_1-f_2=201-200=1pers$