Question

A wave is represented by the equation y = 10 sin 2 pi (100t – 0.02x) + 10 sin 2 pi
(100t + 0.02x). The maximum amplitude and loop length are respectively

Answer 1

Given :

y = 10sin2π (100t - 0.02x) + 10sin2π(100t +0.02x)

To find:

The maximum ampitude and loop length of the wave

Solution :

• y = 10[sin2π (100t - 0.02x) + sin2π(100t +0.02x)]

• y = 10 [2sin2π(100t).sin(0.02x)]

• y = 20[sin2π(100t).sin(0.02x)]

By comparing the above equation with standard equation we get

• Loop length = λ/2

                     = 50 / 2 =25 units

Loop length of wave = 25 units

Maximum amplitude of the wave= 20 units