A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s^{−1} and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the *X*-axis along the string with the origin at one end and *t* = 0 at the instant when the point *x* = 50 cm has reached its maximum displacement.

#### Solution

Given:

Length of the string (*L*) = 2.0 m

Wave speed on the string in its first overtone (*v*) = 200 m/s

Amplitude (*A*) = 0.5 cm

(a) Wavelength and frequency of the string when it is vibrating in its 1^{st} overtone (*n* = 2):

\[L = \frac{n\lambda}{2}\]

\[\Rightarrow \lambda = L = 2 m\]

\[ \Rightarrow f = \frac{\nu}{\lambda} = \frac{200}{2} = 100 \text{ Hz }\]

(b) The stationary wave equation is given by:

\[y = 2A \cos\frac{2\pi x}{\lambda}\sin\frac{2\pi vt}{\lambda}\]

\[ = 0 . 5\cos\frac{2\pi x}{2}\sin\frac{2\pi \times 200 t}{2}\]

\[= \left( 0 . 5 cm \right)\cos\left[ \left( \pi m^{- 1} \right) x \right]\sin\left[ \left( 200\pi s^{- 1} \right) t \right]\]