The first harmonic frequency depends on tension and mass per unit length

Wave Types, Polarisation & Stationary Waves — AQA A-Level Physics

The speed of a wave on a string depends on the tension and the mass per unit length of the string.

v = \sqrt{\frac{T}{\mu}}

$v$: wave speedThe distance travelled by a wavefront per unit time. on the string (m $s^{-1}$)
$T$: tension in the string (N)
$\mu$: mass per unit length (kg $m^{-1}$)

Combining $v = f \lambda$ with $\lambda = 2L$ for the first harmonic gives:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

$f$: frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). of the first harmonic (Hz)
$L$: length of the vibrating string (m)
$T$: tension in the string (N)
$\mu$: mass per unit length (kg $m^{-1}$)

Worked Example

A guitar string has mass 3.2 g and total length 90 cm. It is fixed between two bridges 75 cm apart with a tension of 65 N. Calculate the fundamental frequency.

Show Solution

Calculate mass per unit length

$$\mu = \frac{3.2 \times 10^{-3}}{90 \times 10^{-2}} = 3.56 \times 10^{-3} \text{ kg m}^{-1}$$

Identify the vibrating length

$L = 75 \text{ cm} = 75 \times 10^{-2} \text{ m}$

Note: the vibrating length is between the bridges, not the total string length.

Apply the first harmonic equation

$$f = \frac{1}{2 \times 0.75} \sqrt{\frac{65}{3.56 \times 10^{-3}}} = 90 \text{ Hz (2 s.f.)}$$

Answer

$f = 90$ Hz

Common Mistake MEDIUM

Students often: Using the total string length instead of the vibrating length between the fixed ends.
Instead: L is the distance between the two bridges (or fixed points). The mass per unit length mu uses the total string mass and total string length.