Harmonics of String Oscillation

Classical Mechanics Level 2

An oscillating string of tension $100 \text{ N}$ and mass density per unit length $\mu = 1 \text{ kg}/\text{m}$ fixed at both ends has fundamental frequency $400 \text{ Hz}$ . What is the difference in meters between the wavelengths corresponding to the second and third harmonics?

\frac{1}{400}

\frac{1}{120}

\frac{1}{200}

\frac{1}{240}

1 solution

Matt DeCross
Feb 22, 2016

From applying boundary conditions to standing waves , the frequencies of oscillation go as:

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

Between the second and third harmonics there is therefore a difference of:

$\Delta \lambda = L - \frac{2L}{3} = \frac{L}{3}.$

From the given fundamental frequency,

$\frac{v}{2L} = 400.$

Lastly, since the wave speed is:

$v =\sqrt{\frac{T}{\mu}} = 10$

the length $L$ is $\frac{1}{80}$ , so the difference is $\frac{1}{240}$ as claimed.

Units have been omitted above for brevity.

Harmonics of String Oscillation

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