A wave given with amplitude A = 10 cm travels along a cord which has two sections,one with linear density 0.1 kg/m and the other with linear density 0.2 kg/m.The wave travels along the cord from the lighter density to the heavier.If the tension in the cord is 50 N,the amplitude of the reflected wave is
A)0.65 cm
B)1.5 cm
C)5.9 cm
D)7.4 cm
E)8.3 cm

Question

Quizplus · Accepted Answer

First, we need to find the speed of the wave in each section of the cord. Using the formula:

v = √(T/μ)

where T is the tension in the cord and μ is the linear density, we get:

v1 = √(50/0.1) = 22.36 m/s
v2 = √(50/0.2) = 15.81 m/s

Next, we need to find the reflection coefficient, which is given by:

R = (v2 - v1)/(v2 + v1)

R = (15.81 - 22.36)/(15.81 + 22.36) = -0.209

Since the wave is traveling from the lighter density to the heavier, the reflected wave will be inverted. Therefore, the amplitude of the reflected wave is given by:

Ar = R*A = -0.209*10 = -2.09 cm

But we need to take the absolute value of Ar, since the amplitude cannot be negative:

Ar = 2.09 cm

Therefore, the answer is B) 1.5 cm, since it is the closest to 2.09 cm.

A Wave Given with Amplitude a = 10 Cm Travels