Question

The equation of a wave of amplitude 0.02 m and period 0.04 s travelling along a stretched string with a velocity of 25 m/s will be ____________.

Options

• y = 0.02 sin 2`pi` (0.04t - 0.5x)

• y = 0.02 sin 2`pi` (25t - 2x)

• y = 0.02 sin 2`pi` (0.04t - x)

• y = 0.02 sin 2`pi` (25t - x)

Answer 1

The equation of a wave of amplitude 0.02 m and period 0.04 s travelling along a stretched string with a velocity of 25 m/s will be y = 0.02 sin 2`pi` (25t - x).

Explanation:

`"n" = 1/"T" = 1/0.04 = 25  "Hz"`

v = 25 m/s

`therefore lambda = "v"/"n" = 25/25 = 1  "m"`

Equation of the wave is

`"y" =  "A sin"  2 pi ("t"/"T" - "x"/lambda)`

`therefore "y" = 0.02  "sin"  2 pi  (25"t" - "x")`