One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the breaking stress of the wire is very large and isn’t exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?
Solution 46E Step 1: a) Equation for the frequency, f = v/ Where, v - speed of sound - Wavelength of the sound Provided, the speed of sound in air is, v = 344 m/s Wavelength of the 2nd overtone, = 0.765 m Therefore, frequency of the 2nd overtone, f = 344 m/s / 0.765 m f = 449.67 Hz Step 2: a) equation for the frequency of the sound is, 1 T f = 2L Where, L - length of the string T - Tension on the string - Linear mass density of the string = m/ L Square the equation, 2 1 T f = 4L2 Rearranging the equation to get tension on the string T = 4f L 2 Provided, mass of the string is, m = 8.75 g = 0.00875 kg Length of the string, L = 75 cm = 0.75 m Therefore, = m/L = 0.00875 kg / 0.75 m = 0.012 kg/m Substituting all these values in the above equation for tension we get, 2 2 T = 4 × (449.67 Hz) × (0.75m) × 0.012 kg/m 2 2 T = 4 × 202203 Hz × 0.5625m × 0.012 kg/m T = 5459.5 N Tension acting on the string would be, T = 5459.5 N