A stretched uniform wire of length L under te | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The fundamental frequency of a stretched wire is $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.The fundamental frequency of a closed pipe of length $L$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The fundamental frequency of a stretched wire is $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.The fundamental frequency of a closed pipe of length $L$ is $n = \frac{v}{4L}$.Given that both have the same initial frequency, we have $\frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v}{4L}$.When tension is increased by $16 \,N$, the new frequency $n'$ is $n' = \frac{1}{2L} \sqrt{\frac{T+16}{\mu}}$.This new frequency resonates with the $2^{	ext{nd}}$ harmonic of the closed pipe. The harmonics of a closed pipe are odd multiples of the fundamental frequency $(n, 3n, 5n, \dots)$. However, the question specifies the $2^{	ext{nd}}$ harmonic, which implies the $3n$ frequency (as the $1^{	ext{st}}$ is $n$, $2^{	ext{nd}}$ is $3n$).Thus, $n' = 3n$.$\frac{1}{2L} \sqrt{\frac{T+16}{\mu}} = 3 \left( \frac{1}{2L} \sqrt{\frac{T}{\mu}} ight)$.Squaring both sides: $\frac{T+16}{T} = 3^2 = 9$.$T+16 = 9T \implies 8T = 16 \implies T = 2 \,N$.

Similar Questions

Two tuning forks $A$ and $B$ give $5$ beats per second. Fork $A$ resonates with a closed air column $20 \ cm$ long and fork $B$ with an open air column $40.5 \ cm$ long. Neglecting end correction,the frequencies of tuning forks $A$ and $B$ are respectively.

The length,tension,diameter,and density of a wire $B$ are double those of the corresponding quantities for another stretched wire $A$. Then:

$Assertion :$ The pitch of wind instruments rises and that of string instruments falls as an orchestra warms up.
$Reason :$ When temperature rises,the speed of sound increases,but the speed of a wave in a string fixed at both ends decreases.

Vedclass Test Series

Exam Paper Generator

Online Exam Module