A stretched string of length 0.50\text{ m} fixed at both ends vibrates in its third harmon… — JAMB UTME

A stretched string of length $0.50\text{ m}$ fixed at both ends vibrates in its third harmonic mode at a frequency of $450\text{ Hz}$ . What is the speed of the transverse wave along the string in $\text{m/s}$ ?

Answer: 150 m/s

Answer

The speed of the transverse wave along the string is

150\text{ m/s}

For a string fixed at both ends, standing wave modes produce harmonics given by

f_n = \frac{n v}{2L}

. Given

n = 3

L = 0.50\text{ m}

, and

f_3 = 450\text{ Hz}

, substituting these into the equation yields

450 = \frac{3v}{2(0.50)} = 3v

, leading to

v = 150\text{ m/s}

Step-by-Step Solution

Identify the standing wave frequency equation for a string fixed at both ends.

The frequency of the

n

-th harmonic is

f_n = \frac{n v}{2L}

, where

n

is the harmonic number,

v

is the wave speed, and

L

is the string length.

Fixed ends require nodes at both boundaries, producing standing wave modes with wavelengths

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

Substitute the given physical quantities into the harmonic equation.

450 = \frac{3 \times v}{2 \times 0.50}

The question specifies

L = 0.50\text{ m}

, third harmonic mode (

n = 3

), and frequency

f_3 = 450\text{ Hz}

Solve for the wave speed

v

v = 150\text{ m/s}

Simplifying

2 \times 0.50 = 1.0

gives

3v = 450

, so

v = \frac{450}{3} = 150\text{ m/s}

Key Concept

Standing Waves and Harmonics in Vibrating Strings

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