An organ pipe open at both ends has a length of 0.60\text{ m}. The fundamental frequency o… — JAMB UTME

An organ pipe open at both ends has a length of $0.60\text{ m}$ . The fundamental frequency of resonance for this pipe is observed to be equal to the frequency of the first overtone of a stretched wire of length $0.40\text{ m}$ fixed at both ends. If the linear mass density of the wire is $4.0 \times 10^{-4}\text{ kg/m}$ and the speed of sound in air is $330\text{ m/s}$ , what is the tension in the wire?

A
$1.21\text{ N}$
$4.84\text{ N}$ Answer
C
$9.68\text{ N}$
D
$19.36\text{ N}$

Answer

The tension in the wire is

4.84\text{ N}

The fundamental frequency of the open pipe is

f = \frac{v}{2L} = \frac{330}{2(0.60)} = 275\text{ Hz}

. Since the wire vibrates in its first overtone (second harmonic,

n=2

), its frequency is

f = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{0.40} = 275\text{ Hz}

. This gives

v_{\text{wire}} = 110\text{ m/s}

. Using

v_{\text{wire}} = \sqrt{T/\mu}

, the tension is

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4})(110)^2 = 4.84\text{ N}

Step-by-Step Solution

Calculate the fundamental frequency of the pipe open at both ends.

f_{\text{pipe}} = \frac{v_{\text{air}}}{2 L_{\text{pipe}}} = \frac{330}{2 \times 0.60} = 275\text{ Hz}

An open pipe supports a fundamental wavelength of

\lambda = 2L_{\text{pipe}}

Express the frequency of the first overtone of the stretched wire and equate it to the pipe frequency.

f_{\text{wire, 2}} = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{L_{\text{wire}}} = 275\text{ Hz}

For a wire fixed at both ends, the first overtone corresponds to the second harmonic (

n = 2

Solve for the speed of the transverse wave on the wire.

v_{\text{wire}} = 275 \times 0.40 = 110\text{ m/s}

Rearranging

f = v_{\text{wire}} / L_{\text{wire}}

gives

v_{\text{wire}} = f \cdot L_{\text{wire}}

Calculate the tension in the wire using the wave speed formula.

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4}) \times 110^2 = 4.84\text{ N}

The speed of a transverse wave on a stretched string is given by

v = \sqrt{T / \mu}

Key Concept

Resonance between air columns and vibrating strings under distinct boundary conditions

Estimated Time:2m 0s

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