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Q15.15:
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Solution :

Step 1: Identifying the Physical Principle

For a tube closed at one end (a pipe with one end blocked by a movable piston), resonance occurs at odd harmonics of the fundamental frequency. The resonance condition for length $L$ of a tube open at one end is given by: $L = n\frac{\lambda}{4}$, where $n = 1, 3, 5, \dots$ [Per the theory of standing waves in a pipe closed at one end].

Step 2: Formulating Equations from Observations

Given the two resonant lengths $L_1 = 25.5\text{ cm} = 0.255\text{ m}$ and $L_2 = 79.3\text{ cm} = 0.793\text{ m}$. Since the tuning fork frequency $f = 340\text{ Hz}$ is constant, the distance between two successive resonance positions corresponds to half the wavelength ($\frac{\lambda}{2}$): $\frac{\lambda}{2} = L_2 - L_1$ [Based on the property of successive nodes/antinodes in a pipe].

Step 3: Calculating Wavelength ($\lambda$)

Substituting the given values: $\frac{\lambda}{2} = 0.793\text{ m} - 0.255\text{ m}$ $\frac{\lambda}{2} = 0.538\text{ m}$ $\lambda = 2 \times 0.538\text{ m} = 1.076\text{ m}$

Step 4: Estimating the Speed of Sound ($v$)

The relationship between speed ($v$), frequency ($f$), and wavelength ($\lambda$) is governed by the wave equation $v = f\lambda$ [Fundamental Wave Relation]. $v = 340\text{ Hz} \times 1.076\text{ m}$ $v = 365.84\text{ m/s}$

Step 5: Visual Representation of Resonance

Resonant Air Column Open End Piston Length L

Final Solution: The speed of sound in air at the temperature of the experiment is $365.84\text{ m/s}$.


More Questions from Class 11 Physics Waves EXERCISES


CBSE Solutions for Class 11 Physics Waves


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