A steel wire with a mass of 4.0 g and a length of 80 cm is fixed at both its ends. The tension in the wire is 50 N . Determine the frequency and wavelength of the fourth harmonic of the fundamental.

Correct answer: A. 250 Hz , 40 cm Linear mass density of the wire is μ = (m)/(L) = 4.0 × 10^ -3 0.8 = 5.0 × 10^ -3 kg/m The speed of the transverse wave on the string is v = (T)/(μ) = 50 5.0 × 10^ -3 = 100 m/s For a string fixed at both ends, the frequency of the n -th harmonic is given by f_n = (nv)/(2L) For the fourth harmonic, n = 4 f_4 = (4 × 100)/(2 × 0.8) = 250 Hz The wavelength of the n -th harmonic is given by λ_n = (2L)/(n) λ_4 = (2 × 80)/(4) = 40 cm

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A steel wire with a mass of 4.0 g and a length of 80 cm is fixed at...

A steel wire with a mass of

4.0 \text{ g}

and a length of

80 \text{ cm}

is fixed at both its ends. The tension in the wire is

50 \text{ N}

. Determine the frequency and wavelength of the fourth harmonic of the fundamental.

Options

A $250 \text{ Hz}$ , $40 \text{ cm}$
B $62.5 \text{ Hz}$ , $160 \text{ cm}$
C $500 \text{ Hz}$ , $20 \text{ cm}$
D $125 \text{ Hz}$ , $80 \text{ cm}$

Correct answer

A. $250 \text{ Hz}$ , $40 \text{ cm}$

Step-by-step solution

Linear mass density of the wire is

\mu = \dfrac{m}{L} = \dfrac{4.0 \times 10^{-3}}{0.8} = 5.0 \times 10^{-3} \text{ kg/m}

The speed of the transverse wave on the string is

v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{50}{5.0 \times 10^{-3}}} = 100 \text{ m/s}

For a string fixed at both ends, the frequency of the

n

-th harmonic is given by

f_n = \dfrac{nv}{2L}

For the fourth harmonic,

n = 4

f_4 = \dfrac{4 \times 100}{2 \times 0.8} = 250 \text{ Hz}

The wavelength of the

n

-th harmonic is given by

\lambda_n = \dfrac{2L}{n}

\lambda_4 = \dfrac{2 \times 80}{4} = 40 \text{ cm}

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