Question

A string with a length of 1.5 m resonates in three loops as shown above. The string linear density is 0.03 kg/m and the suspended mass is 1.2 kg.

a.What is the wavelength?

b.What is the wave speed?

c.What is the frequency of oscillations?

d.What will happen to the number of loops if the suspended mass is increased?

Answer 1

Answer:

Given that, length of the string L = 1.5 m, linear density = 0.03 kg/m, mass m = 1.2 kg.and number of loops n = 3.

Tension in he string T = mg = (1.2 kg) (9.8 m/s²) = 11.77 N

(a) Wavelength = 2L = 2(1.5 m) = 3.0 m.

(b) Speed of wave in the string v = (T/)^1/2 = (11.77 N/0.03 kg/m)^1/2 = 19.8 m/s

(c) Frequency of oscillation f = nv/2L = (3)(19.8 m/s) / 2(1.5 m) = 19.8 Hz

(d) If the suspended mass is increased then the tension in the string will increase, due to increase in the tension the speed of the wave increase.Therefore, due to the increase in the speed, the number of loops (n) in the string will decrease.

Since speed v = 2Lf/n. i.e., v is inversely proportional to n.