A stretched string of length 0.5\text{ m} fixed at both ends vibrates in its fundamental m… — JAMB UTME

A stretched string of length $0.5\text{ m}$ fixed at both ends vibrates in its fundamental mode. If the speed of transverse waves along the string is $200\text{ m/s}$ , calculate the fundamental frequency of the string in hertz.

Answer: 200 Hz

Answer

The fundamental frequency of the vibrating string is

200\text{ Hz}

For a string fixed at both ends, the fundamental mode corresponds to a standing wave with half a wavelength spanning the length of the string (

L = \frac{\lambda}{2}

, or

\lambda = 2L

). Applying the wave relation

v = f\lambda

, the fundamental frequency is

f = \frac{v}{2L}

. Substituting

v = 200\text{ m/s}

and

L = 0.5\text{ m}

gives

f = \frac{200}{2(0.5)} = 200\text{ Hz}

Step-by-Step Solution

Identify the relationship between frequency, wave speed, and string length for the fundamental mode.

For a string fixed at both ends, the wavelength of the fundamental harmonic is

\lambda = 2L

, giving the frequency formula

f = \frac{v}{2L}

The fundamental standing wave pattern contains nodes at both fixed ends and a single antinode at the center.

Substitute the given numerical values into the formula.

f = \frac{200\text{ m/s}}{2 \times 0.5\text{ m}} = \frac{200}{1} = 200\text{ Hz}

Dividing the wave speed by twice the length of the string yields the frequency in hertz.

Key Concept

Fundamental frequency of a vibrating string fixed at both ends

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