A simple harmonic oscillator at the point x = 0 generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function y(x, t) for the wave. Assume that the oscillator has its maximum upward displacement at time t = 0. (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

Solution 21E Introduction The frequency of the wave will be equal to the frequency of the oscillator. Now knowing the tension of the string and density we can calculate the speed of the wave on the rope. Now from the speed of the wave and frequency we can calculate the wavelength and hence the wavenumber. Then, using the initial condition we can write the wave equation. From the wave equation we can calculate maximum transverse acceleration. Then we have to discuss if it is okay to ignore the gravitational acceleration. Step 1 The linear mass density of the rope is = 50.0 g/m = 50.0 kg/m And the tension on the string is = 5.00 N Hence the speed of the wave is given by Step 2 The frequency of the wave will be equal to the frequency of the oscillator, hence the frequency of the wave is f = 40.0 Hz . Hence we the wavelength is given by Step 3 The wave function of the wave can be written as Now we have And Also as given at t = 0, and x = 0 we have Hence we have Hence the wave equation becomes