6. ⇒ (MHT CET 2023 9th May Evening Shift)

If the length of an open organ pipe is $33.3 cm$ , then the frequency of fifth overtone is [Neglect end correction, velocity of sound $= 333 m / s$ ]

A. $3500 Hz$

B. $3000 Hz$

C. $2500 Hz$

D. $2000 Hz$

Correct Option is (B)

For a pipe open at both ends,

$n = \frac{v}{21} = \frac{333}{2 \times 33.3 \times 10^{- 2}} = 500 Hz$

$∴$ Frequency of $5^{th}$ overtone,

$n = 6 n = 6 \times 500 = 3000 Hz$

7. ⇒ (MHT CET 2023 9th May Evening Shift )

If the end correction of an open pipe is $0.8 cm$ , then the inner radius of that pipe is

A. $\frac{1}{3} cm$

B. $\frac{2}{3} cm$

C. $\frac{3}{2} cm$

D. $0.2 cm$

Correct Option is (B)

For an open pipe, $e = 0.6 d$

$\begin{array}{ll} ∴ & d = \frac{e}{0.6} \\ ∴ & 2 r = \frac{e}{0.6} \\ ∴ & r = \frac{0.8}{1.2} = \frac{2}{3} cm \end{array}$

8. ⇒ (MHT CET 2023 9th May Morning Shift )

A closed organ pipe of length ' $L_{1}$ ' and an open organ pipe contain diatomic gases of densities ' $ρ_{1}$ ' and ' $ρ_{2}$ ' respectively. The compressibilities of the gases are same in both pipes, which are vibrating in their first overtone with same frequency. The length of the open organ pipe is (Neglect end correction)

A. $\frac{4 L_{1}}{3}$

B. $\frac{4 L_{1}}{3} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

C. $\frac{4 L_{1}}{3} \sqrt{\frac{ρ_{2}}{ρ_{1}}}$

D. $\frac{3}{4 L_{1}} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

Correct Option is (B)

Given both gases are vibrating in the first overtone with same frequency, we get

$\begin{aligned} f_{closed} = f_{open} \\ \Rightarrow \frac{3 v}{4 L_{1}} = \frac{v}{L_{2}} \end{aligned}$

According to Laplace's correction

$\begin{aligned} v = \sqrt{\frac{γ P}{ρ}} \\ \frac{3}{4 L_{1}} \times \sqrt{\frac{γ P}{ρ_{1}}} = \frac{1}{L_{2}} \times \sqrt{\frac{γ P}{ρ_{2}}} \\ ∴ L_{2} & = \frac{4 L_{1}}{3} \sqrt{\frac{ρ_{1}}{ρ_{2}}} \end{aligned}$

9. ⇒ (MHT CET 2021 21th September Evening Shift )

A pipe closed at one end has length $0.8 m$ . At its open end $0.5 m$ long uniform string is vibrating in its $2^{nd}$ harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 N$ and the speed of sound is $320 m / s$ , the mass of the string is

A. 20 gram

B. 10 gram

C. 5 gram

D. 15 gram

Correct Option is (B)

$\begin{aligned} 2 \times [\frac{1}{2 ℓ_{1}} \sqrt{\frac{T}{m}}] = \frac{v}{4 ℓ_{2}} \\ \frac{1}{0.5} \sqrt{\frac{50}{m}} = \frac{320}{4 \times 0.8} \\ ∴ m = 0.02 kg / m \\ ∴ Total mass of the string \\ = 0.02 \times 0.5 kg = 10 gm \end{aligned}$

10. ⇒ (MHT CET 2021 21th September Morning Shift )

A closed organ pipe of length ' $L_{c}$ ' and an open organ pipe of length ' $L_{o}$ ' contain different gases of densities ' $ρ_{1}$ ' and ' $ρ_{2}$ ' respectively. The compressibility of the gases is the same in both the pipes. The gases are vibrating in their first overtone with the same frequency. What is the length of open organ pipe?

A. $\frac{4 L_{c}}{3} \sqrt{\frac{ρ}{ρ_{2}}}$

B. $\frac{3 L_{c}}{4} \sqrt{\frac{ρ_{2}}{ρ_{1}}}$

C. $\frac{4 L_{c}}{3} \sqrt{\frac{ρ_{2}}{ρ_{1}}}$

D. $\frac{2 L_{c}}{3} \sqrt{\frac{ρ_{2}}{ρ}}$

Correct Option is (A)

For open organ pipe :

Fundamental frequency $n = \frac{V}{2 L_{o}}$

First overtone $n_{1} = 2 n = \frac{V}{L_{o}}$

For closed organ pipe

Fundamental frequency $n^{'} = \frac{V^{'}}{4 L_{c}}$

First overtone $n_{1}^{'} = 3 n^{'} = \frac{3 V^{'}}{4 L_{c}}$

$\begin{aligned} n_{1} & = n_{1}^{'} ∴ \frac{V}{L_{o}} = \frac{3 V^{'}}{4 L_{c}} \\ ∴ \frac{L_{0}}{L_{c}} & = \frac{4}{3} \frac{V}{V^{'}} .......(1) \\ V & = \sqrt{\frac{k}{ρ_{2}}}, V^{'} = \sqrt{\frac{k}{ρ_{l}}} \end{aligned}$

where $k$ is the adiabatic bulk modulus, which is reciprocal of compressibility.

$∴ \frac{V}{V^{'}} = \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

Putting this value in Eq.(1)

$\frac{L_{0}}{L_{c}} = \frac{4}{3} \sqrt{\frac{ρ_{1}}{ρ_{2}}}$

⇒Superposition of Waves