A sonometer wire of length 0.80\text{ m} and mass 2.0\text{ g} is maintained under a tensi… — JAMB UTME

A sonometer wire of length $0.80\text{ m}$ and mass $2.0\text{ g}$ is maintained under a tension of $100\text{ N}$ . What is the fundamental frequency of vibration of the wire?

$125\text{ Hz}$ Answer
B
$250\text{ Hz}$
C
$62.5\text{ Hz}$
D
$12.5\text{ Hz}$

Answer

The fundamental frequency of vibration of the wire is

125\text{ Hz}

The linear mass density is

\mu = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m}

. The velocity of waves on the string is

v = \sqrt{\frac{100}{0.0025}} = 200\text{ m/s}

. The fundamental frequency is

f = \frac{v}{2L} = \frac{200}{2 \times 0.80} = 125\text{ Hz}

Step-by-Step Solution

Convert mass to kilograms and calculate linear mass density

(\mu)

m = 2.0\text{ g} = 0.002\text{ kg}

. Thus,

\mu = \frac{m}{L} = \frac{0.002\text{ kg}}{0.80\text{ m}} = 0.0025\text{ kg/m} = 2.5 \times 10^{-3}\text{ kg/m}

Standard SI units must be used for tension in Newtons and length in meters.

Calculate the speed of the transverse wave on the string

(v)

v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.0025}} = \sqrt{40000} = 200\text{ m/s}

Wave speed on a stretched string depends directly on tension and inversely on linear density.

Calculate the fundamental frequency

(f_1)

f_1 = \frac{v}{2L} = \frac{200}{2 \times 0.80} = \frac{200}{1.6} = 125\text{ Hz}

For a fixed string vibrating in its fundamental mode, the length equals half the wavelength (

\,L = \frac{\lambda}{2}\,

Key Concept

Fundamental frequency of a stretched string

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