Question

One end of a horizontal string is attached to a small-amplitude mechanical 60.0 Hz oscillator. The string's mass per unit length is 3.9 x 10^{-4} kg/m. The string passes over a pulley, a distance L = 1.50 m away, and weights are hung from this end. Assume the string at the oscillator is a node, which is nearly true.

 

A. What mass m must be hung from this end of the string to produce one loop of a standing wave?

B. What mass m must be hung from this end of the string to produce two loops of a standing wave?

C. What mass m must be hung from this end of the string to produce five loops of a standing wave?

 

Answer 1

string's mass per unit length = 3.9 x 10^{-4} kg/m.

60.0 Hz oscillator.

a)one loop in kg

L = 1.50 one loop means wavelength should be 2L= 3.0m; frequency = 60

so velocity = 3*60 = 180 m/s

We have 180 = sqrt[T/m],

where T is tension in string T= M*g with M kg on the pan So we have

M = (180^2)*(m/g)

M = [m*(180^2)]/g

= [(3.9*10^-4)*180^2]/9.8 = 1.289kg =1.29 Kg

b)two loops in kg

wavelength = L = 1.50 m or V = 60*1.50 = 90;

So M is given by

M = [(3.9*10^-4)*(90)^2]/9.8 = 0.322 kg = 322 grams

c)five loops in kg

L = 1.50 m (wavelength/2)*5 = 1.50 or wavelength = 3/5 = 0.6 m

Velocity = 60*0.6 = 36 m/s

or M = [(3.9*10^-4)*(36)^2]/9.8 = 0.0515 kg = 51 grams