Question 305: Waves

Question

A wire of length ' $L$ ' and linear density ' $m$ ' is stretched between two rigid supports with tension ' $T$ '. It is observed that wire resonates in the $\mathrm{P}^{	ext {th }}$ harmonic at a frequency of 320 Hz and resonates again at next higher frequency of 400 Hz . The value of ' $p$ ' is

Accepted Answer

We are given that the frequencies of the harmonics on a stretched wire are given by:
$$ f_n = \frac{n}{2L} \sqrt{\frac{T}{m}} $$
where:
Ã¢ÂÂ¢ $$ n $$ represents the harmonic number,
Ã¢ÂÂ¢ $$ L $$ is the length,
Ã¢ÂÂ¢ $$ T $$ is the tension,
Ã¢ÂÂ¢ $$ m $$ is the linear density.
LetÃ¢ÂÂs denote:
$$ k = \frac{1}{2L} \sqrt{\frac{T}{m}} $$
Then:
Ã¢ÂÂ¢ The frequency of the $$ p^{	ext{th}} $$ harmonic is:
$$ f_p = p \cdot k = 320\ 	ext{Hz} $$
Ã¢ÂÂ¢ The frequency of the next higher harmonic (i.e., $$ (p+1)^{	ext{th}} $$) is:
$$ f_{p+1} = (p+1) \cdot k = 400\ 	ext{Hz} $$
We can form the ratio of these two equations to eliminate $$ k $$:
$$ \frac{f_{p+1}}{f_p} = \frac{(p+1)k}{p k} = \frac{p+1}{p} = \frac{400}{320} = \frac{5}{4} $$
Now, solve for $$ p $$:
Write the equation:
$$ \frac{p+1}{p} = \frac{5}{4} $$
Cross multiply:
$$ 4(p + 1) = 5p $$
Simplify:
$$ 4p + 4 = 5p $$
Solve for $$ p $$:
$$ 5p - 4p = 4 $$
$$ p = 4 $$
Thus, the value of $$ p $$ is 4.
The correct answer is Option B.

Answer

2

Answer

4

Answer

8

Answer

10

Waves

A wire of length ' L ' and linear density ' m ' is stretched between two rigid supports with tension ' T '. It is observed that wire resonates in the \mathrm{P}^{\text {th }} harmonic at a frequency of 320 Hz and resonates again at next higher frequency of 400 Hz . The value of ' p ' is

Choose the correct answer: