Question

Two identical strings of the same material, same diameter and same length are in unison, when stretched by the same tension. If the tension on one string is increased by $21$%, the number of beats heard per second is $10$. The frequency of the note in hertz, when the strings are in unison is :

• A. $210$
• B. $200$
• C. $110$
• D. $100$

Answer 1

The correct option is D $100$

For transverse vibration in the string, frequency is given by
$n=\frac{1}{2l}\sqrt{\frac{T}{m}}$
(where T is tension, m is the mass per unit length)
Hence, $n_1=\frac{1}{2l}\sqrt{\frac{T_1}{m}}..\left(i\right)$
and $n_2=\frac{1}{2l}\sqrt{\frac{T_2}{m}}..\left(ii\right)$
Number of beat $n_2-n_1=10...\left(iii\right)$
Given $T_2=T_1+\frac{21}{100}{T}_1=1.21T_1$
Putting the value of $T_2$ in Eq. (ii), we get
$n_2=\frac{1}{2I}\sqrt{1.21\frac{T_1}{m}}=1.1\times \frac{1}{2l}\sqrt{\frac{T_1}{m}}$
$n_2=1.1n_1...\left(v\right)$
Putting the value of $n_2$ from Eq. (v) in Eq. (iii), we get;
$1.1n_1-n_1=10$
$0.1n_1=10$
$n_1=100Hz$