Q.A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\ \text{Hz}$. The mass of the wire is $3.5 \times 10^{-2}\ \text{kg}$ and its linear mass density is $4.0 \times 10^{-2}\ \text{kg m}^{-1}$. What is

Concept: Wave speed on a stretched string and fundamental-mode vibration.

The wire vibrates in its fundamental mode, so the length $L$ equals half a wavelength: $L = \frac{\lambda}{2}$, giving $\lambda = 2L$.

First find the length from the given mass and linear density:

(a) The wave speed relates to frequency and wavelength by $v = f\lambda$:

$v = f \cdot 2L = 45 \times 2 \times 0.875 = 78.75\ \text{m s}^{-1}$ …

Understanding Wave Speed and Tension on a Vibrating String

When a string fixed at both ends vibrates, it supports standing waves. The fundamental mode (first harmonic) is the simplest pattern: the string forms a single "hump" with nodes at the two fixed ends and one antinode in the middle. The wavelength of this fundamental mode is exactly twice the length of the string, $\lambda = 2L$.

The speed of a transverse wave on any string depends on two physical properties: the tension $T$ pulling the string taut, and the linear mass density $\mu$ (mass per unit length). These combine in the beautiful relation

This tells us that waves travel faster on tighter, lighter strings—exactly what a guitarist knows intuitively.

For a standing wave, the relationship between wave speed, frequency, and wavelength is the standard wave equation $v = f\lambda$. Once we know the length of the wire, we can find the wavelength of the fundamental mode, then the speed, and finally back out the tension.

Solution

1. Find the length of the wire

We're given the total mass $m = 3.5 \times 10^{-2} \text{ kg}$ and the linear mass density $\mu = 4.0 \times 10^{-2} \text{ kg/m}$. Since $\mu = m/L$, we have

2. Determine the wavelength in fundamental mode

For the fundamental mode (first harmonic), the string vibrates with one antinode. The wavelength is

3. Calculate the wave speed

The frequency is given as $f = 45 \text{ Hz}$. Using the wave equation:

$v = f\lambda = 45 \times 1.75 = 78.75 \text{ m/s}$ …

Step 1: For a string fixed at both ends vibrating in its fundamental mode, the length equals half a wavelength: $L=\lambda/2$.

Step 2: Find $L$ from the given mass and linear density: $L=m/\mu=(3.5\times10^{-2})/(4.0\times10^{-2})=0.875\ \text{m}$, so $\lambda=2L=1.75\ \text{m}$. …