A uniform string is fixed at one end and passes over a smooth pulley to support a hanging block of mass M . The length of the string between the fixed end and the pulley has a mass of 10 g and a linear mass density of 5 g m ⁻¹ . When this segment of the string vibrates in its third harmonic, the frequency is 150 Hz . Taking g = 10 m s ⁻² , the value of M is _______ kg .

Step-by-step solution

The length of the vibrating segment of the string is: l = m = 10 g 5 g m ⁻¹ = 2 m For the third harmonic, the frequency is given by: f = 3v 2l Substituting the given values: 150 = 3v 2 2 v = 150 4 3 = 200 m s ⁻¹ The linear mass density in SI units is = 5 g m ⁻¹ = 0.005 kg m ⁻¹ . The speed of a transverse wave is v = T , so the tension T in the string is: T = v^2 = (200)^2 0.005 = 40000 0.005 = 200 N This tension is provided by the weight of the hanging block: T = Mg 200 = M 10 M = 20 kg Answer: 20