Question

Transverse waves are being generated on a rope under constant tension. By what factor is the required power increased or decreased if (a) the length of the rope is doubled and the angular frequency remains constant, (b) the amplitude is doubled and the angular frequency is halved, (c) both the wavelength and the amplitude are doubled, and (d) both the length of the rope and the wavelength are halved?

Answer 1

Answer:

a) when the length of the rope is doubled and the angular frequency remains constant: The power increases by factor of two

b) when the amplitude is doubled and the angular frequency is halved: The power is the same

c) when both the wavelength and the amplitude are doubled: The power increases by a factor of 8

d) when both the length of the rope and the wavelength are halved: The power increases by factor of two

Explanation:

For a sinusoidal mechanical wave (Transverse wave), the time-averaged power is the energy associated with a wavelength divided by the period of the wave.

$P_{avg} =\frac{E \lambda}{T}=\frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T}$

where;

A is the Amplitude

ω is the angular frequency

λ is the wavelength

a) when the length of the rope is doubled and the angular frequency remains constant

L = λ/2,  λ = 2L

The power increases by factor of two

b) when the amplitude is doubled and the angular frequency is halved

$P_{avg} ={\frac{1}{2} \mu (2A)^2 (\frac{\omega}{2}) ^2 \frac{\lambda}{T} = \frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T}$

The power is the same

c) when both the wavelength and the amplitude are doubled

$P_{avg} =\frac{1}{2} \mu (2A)^2 \omega^2 \frac{(2\lambda)}{T} = \frac{1}{2} \mu (4A^2) \omega^2 \frac{(2\lambda)}{T} = 8 (\frac{1}{2} \mu A^2 \omega^2 \frac{\lambda}{T})$

The power increases by a factor of 8

d) when both the length of the rope and the wavelength are halved

L = λ/2

when both are halved

L/2 =  λ/4, λ = 2L

The power increases by factor of two