Question

The arrangement in the drawing shows a block (mass = 15.0 kg) that is held in position on a frictionless incline by a cord (length = 0.52 m). The mass per unit length of the cord is 1.26 × 10-2 kg/m, so the mass of the cord is negligible compared to the mass of the block. The cord is being vibrated at a frequency of 91.7 Hz (vibration source not shown in the drawing). What is the largest angle θ at which a standing wave exists on the cord?

Answer 1

the tension in the cord must equal the component of the block's weight down the plane; this tension will equal

T = m g sin(theta) where m is the mass of the block

c = Sqrt[T/u] where T is the tension and u is the mass density of the string (given as 0.0126 kg/m)

frequency of the resonances on a string is dependent on the speed of sound in the string

f = c/2L where L is the length of the string

the allowable frequencies are integral multiples of the fundamental, so we have

f = n c/2L or c = 2 L f/n



we know that c = Sqrt[T/u] and that T = mg sin(theta)

substitute this expression for T and c = 2Lf/n and get

2 L f/n = Sqrt[ m g sin(theta)/u]

square both sides and solve for sin(theta)

sin(theta) = 4 L^2 f^2 u/(m g n^2)

Put the value of L=0.52m, f = 91.7Hz, u = 0.0126kg/m, m = 15.0kg and g = 9.81 m/s/s, we get

0.778/n^2 = sin(theta)

n is an integer, so we can see immediately that n=1

theta = 51.07 deg