21. (JEE Main 2018 (Online) 15th April Evening Slot)

5 beats / econd are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :

A. 195 Hz

B. 150 Hz

C. 300 Hz

D. 251 Hz

Correct Answer is Option (A)

Length of wire is L₁ = 0.95 m; L₂ = 1 m. Number of beats per second heard = 5

Let frequency of fork be f. Therefore,

$\frac{v}{2 L_{1}} - f = 5 \Rightarrow \frac{v}{2 L_{1}} = 5 + f$ ...... (1)

and $f - \frac{v}{2 L_{2}} = 5 \Rightarrow \frac{v}{2 L_{2}} = f - 5$ .... (2)

Dividing Eq. (1) by Eq. (2), we get

$\frac{\frac{v}{2 L_{1}}}{\frac{v}{2 L_{2}}} = \frac{5 + f}{f - 5} \Rightarrow \frac{L_{2}}{L_{1}} = \frac{f + 5}{f - 5}$

$\Rightarrow \frac{1}{0.95} = \frac{f + 5}{f - 5} \Rightarrow f - 5 = 0.95 f + 4.75$

$\Rightarrow f - 0.95 f = 5 + 4.75 \Rightarrow 0.05 f = 9.75 \Rightarrow f = \frac{9.75}{0.05}$

$\Rightarrow f = 195$ Hz

22. (JEE Main 2017 (Online) 8th April Morning Slot)

Two wires W₁ and W₂ have the same radius r and respective densities $ρ$ ₁ and $ρ$ ₂ such that ρ₂ = 4 $ρ$ ₁ . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W₁ to W₂ is :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Waves Question 93 English

A. 1 : 1

B. 1 : 2

C. 1 : 3

D. 4 : 1

Correct Answer is Option (B)

Fundamental frequency of a stretched string is ,

f = $\frac{n}{2 L} \sqrt{\frac{T}{μ}}$

Here, n = number of antinodes

$μ$ = mass per unit length.

As, O is the midpoint of two bridegs, hence length of two wires are equal.

$∴$ L₁ = L₂ = L

As frequency of both wires same

f₁ = f₂

$\Rightarrow$ $\frac{n_{1}}{2 L} \sqrt{\frac{T}{π r^{2} ρ_{1}}}$ = $\frac{n_{2}}{2 L} \sqrt{\frac{T}{π r^{2} ρ_{2}}}$

$\Rightarrow$ $\frac{n_{1}}{n_{2}}$ = $\sqrt{\frac{ρ_{1}}{ρ_{2}}}$

$\Rightarrow$ $\frac{n_{1}}{n_{2}}$ = $\sqrt{\frac{ρ_{1}}{4 ρ_{1}}}$ = $\frac{1}{2}$

23. (JEE Main 2016 (Offline))

A uniform string of length $20$ $m$ is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the supports is :
(take $g = 10 m s^{- 2}$ )

A. $2 \sqrt{2} s$

B. $2 π \sqrt{2} s$

C. $2 π \sqrt{2} s$

D. 2 s

Correct Answer is Option (A)

JEE Main 2016 (Offline) Physics - Waves Question 101 English Explanation
We know that velocity in string is given by

$v = \sqrt{\frac{T}{μ}} . . . . (i)$

where $μ = \frac{m}{I} = \frac{m a s s o f s t r i n g}{l e n g t h o f s t r i n g}$

The tension $T = \frac{m}{ℓ} \times x \times g . . . (i i)$

From $(a)$ and $(b)$ $\frac{d x}{d t} = \sqrt{g x}$

$x^{- 1 / 2} d x = \sqrt{g} d t$

$∴$ $\int_{0}^{ℓ} x^{- 1 / 2} d x - \sqrt{g} \int_{0}^{ℓ} d t$

$2 \sqrt{ℓ} = \sqrt{g} \times t$

$∴$ $t = 2 \sqrt{\frac{ℓ}{g}} = 2 \sqrt{\frac{20}{10}} = 2 \sqrt{2}$

24. (JEE Main 2013 (Offline))

A sonometer wire of length $1.5$ $m$ is made of steel. The tension in it produces an elastic strain of $1 %$ . What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^{3} k g / m^{3}$ and $2.2 \times 10^{11} N / m^{2}$ respectively ?

A.

B.

C.

D.

Correct Answer is Option ()

25. (AIEEE 2010)

The equation of a wave on a string of linear mass density $0.04 k g m^{- 1}$ is given by $y = 0.02 (m) \sin [2 π (\frac{t}{0.04 (s)} - \frac{x}{0.50 (m)})] .$

The tension in the string is

A. 4.0 N

B. 12.5 N

C. 0.5 N

D. 6.25 N

Correct Answer is Option (D)

$y = 0.02 (m) \sin [2 π (\frac{t}{0.04 (s)}) - \frac{x}{0.50 (m)}]$

But $y = a \sin (ω t - k x)$

$∴$ $ω = \frac{2 π}{0.04} \Rightarrow v = \frac{1}{0.04} = 25 H z$

$k = \frac{2 π}{0.50} \Rightarrow λ = 0.5 m$

$∴$ velocity, $v = v λ = 25 \times 0.5 m / s = 12.5 m / s$

Velocity on a string is given by

$v = \sqrt{\frac{T}{μ}}$

$∴$ $T = v^{2} \times μ = {(12.5)}^{2} \times 0.04 = 6.25 N$