A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50N and the speed of sound is 320 ms

Engineering

Physics

Standing Wave on String

Quinkes Kundts and Resonance Tube Experiment

Interference of Sound Wave

Question

A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50N and the speed of sound is 320 ms^–1, the mass of the string is

40 grams

20 grams

10 grams

5 grams

$\frac{2}{2 \times 0 .5} \times \sqrt{\frac{\frac{50}{m}}{0 .5}} = \frac{1}{4 \times 0 .8} \times 320$

$\frac{2 \times 5}{\sqrt{m}} - 100$

$m = \frac{1}{100} = 10 gm$

This problem involves resonance between a vibrating string and an air column in a pipe. Let's break it down step by step.

Step 1: Understand the Systems

We have two systems:

A hollow pipe, closed at one end, of length L_p = 0.8 m.
A uniform string, of length L_s = 0.5 m, vibrating in its second harmonic.

They resonate, meaning their frequencies are equal. The string's second harmonic frequency equals the pipe's fundamental frequency.

Step 2: Find the Fundamental Frequency of the Closed Pipe

For a pipe closed at one end, the fundamental frequency (first harmonic) is given by:

f_{p} = \frac{v}{4 L_{p}}

where v = 320 m/s (speed of sound) and L_p = 0.8 m.

Let's calculate this:

f_{p} = \frac{320}{4 \times 0.8} = \frac{320}{3.2} = 100 Hz

So, the fundamental frequency of the pipe is 100 Hz.

Step 3: Relate to the String's Second Harmonic

The string is vibrating in its second harmonic. The frequency of the n^th harmonic for a string fixed at both ends is:

f_{n} = \frac{n}{2 L_{s}} \sqrt{\frac{T}{μ}}

where:

n = harmonic number (n=2 for second harmonic)
L_s = length of the string = 0.5 m
T = Tension in the string = 50 N
μ = mass per unit length of the string (which we need to find)

Since this frequency (f₂) resonates with the pipe's frequency, we set them equal:

f_{2} = f_{p} = 100 Hz

Therefore, we can write the equation:

\frac{2}{2 L_{s}} \sqrt{\frac{T}{μ}} = 100

This simplifies to:

\frac{1}{L_{s}} \sqrt{\frac{T}{μ}} = 100

Step 4: Solve for Mass per Unit Length (μ)

Let's plug in the known values (L_s = 0.5 m, T = 50 N) and solve for μ.

\frac{1}{0.5} \sqrt{\frac{50}{μ}} = 100

2 \sqrt{\frac{50}{μ}} = 100

\sqrt{\frac{50}{μ}} = 50

Now, square both sides to eliminate the square root:

\frac{50}{μ} = 2500

μ = \frac{50}{2500} = 0.02 kg/m

So, the mass per unit length of the string is 0.02 kg/m.

Step 5: Find the Total Mass of the String

The total mass (m) of the string is its mass per unit length (μ) multiplied by its total length (L_s).

m = μ \times L_{s} = 0.02 kg/m \times 0.5 m = 0.01 kg

To convert this mass into grams, we multiply by 1000:

m = 0.01 \times 1000 = 10 grams

Final Answer

The mass of the string is 10 grams.

Column-I	Column-II
(A) Pipe closed at one end	(p) Longitudinal waves
(B) Pipe open at both ends	(q) Transverse waves
(C) Stretched wire clamped at both ends	(r) $λ$ _f = L
(D) Stretched wire clamped at both ends and at mid-point.	(s) $λ$ _f = 2L
	(t) $λ$ _f = 4L

Detailed Solution

Step 1: Understand the Systems

Step 2: Find the Fundamental Frequency of the Closed Pipe

Step 3: Relate to the String's Second Harmonic

Step 4: Solve for Mass per Unit Length (μ)

Step 5: Find the Total Mass of the String

Final Answer

Related Topics & Formulae

Standing Waves in Strings

Standing Waves in Air Columns

Resonance

Detailed Solution

Step 1: Understand the Systems

Step 2: Find the Fundamental Frequency of the Closed Pipe

Step 3: Relate to the String's Second Harmonic

Step 4: Solve for Mass per Unit Length (μ)

Step 5: Find the Total Mass of the String

Final Answer

Related Topics & Formulae

Standing Waves in Strings

Standing Waves in Air Columns

Resonance

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