Question

Which of the followings represent the variation of frequency (f) with radius of cross section of a stretched string?

Answer 1

Answer:

Frequency is inversely proportional to the radius of cross section of a stretched string.

Answer 2

For the whole question find the attached photo.

Answer:

Option (A) is the correct answer.

Explanation:

Step 1: $Frequency (f) = \frac{1}{2l} \sqrt{ \frac{T}{m} }$

Where, T = tension

m = mass per length

Step 2: But, m = mass / length of wire

So,

$mass \: = \: \frac{volume \: \times density}{legth \: of \: wire}$

Step 3: But, volume = area × length

So,

$mass \: = \frac{area \: \times \: length \: \times density}{length}$

Step 4: Length / length will cancel and we get,

$mass \: = area (πr²) \: \times density (ρ)$

So, m = πr² × ρ

Step 5: Putting the above equation in the frequency equation,

$f = \frac{1}{2l} \sqrt{ \frac{T}{πr²ρ} }$

Step 6: But,

$\sqrt{ \frac{1}{ {r}^{2} } } = \frac{1}{r}$

So,

$f = \frac{1}{2lr} \sqrt{ \frac{T}{πρ} }$

Step 7: But other than r, every variable is constant.

So, frequency (f) is inversely proportional to r and give hyper bolic graph.

So, option (A) is the correct answer.