Question:

A wave traveling on a string has a wavelength of 0.1 m and a frequency of 100 Hz. The wave propagates with a velocity of 5 m/s.

a) Determine the amplitude of the wave.

b) Calculate the period and angular frequency of the wave.

c) Find the maximum displacement and speed of a point on the string.

Answer:

a) The amplitude of a wave is the maximum displacement of particles from their equilibrium position. It can be determined using the equation:

Amplitude (A) = maximum displacement / 2

However, in this question, the maximum displacement is not provided directly. Instead, we can calculate it using the wave equation:

Velocity (v) = frequency (f) × wavelength (λ)

Rearranging the equation to solve for the maximum displacement:

Maximum displacement = Velocity / frequency

Substituting the given values:

Maximum displacement = 5 m/s / 100 Hz = 0.05 m

Therefore, the amplitude of the wave is 0.05 m.

b) The period (T) of a wave is the time it takes for one complete oscillation. It is the reciprocal of the frequency, so:

Period (T) = 1 / frequency (f)

Substituting the given frequency:

Period (T) = 1 / 100 Hz = 0.01 s

The angular frequency (ω) is a measure of how quickly a wave oscillates. It is expressed in radians per second and is related to the period by the equation:

Angular frequency (ω) = 2π / Period (T)

Substituting the value for the period:

Angular frequency (ω) = 2π / 0.01 s = 200π rad/s (approximately)

Therefore, the period of the wave is 0.01 seconds and the angular frequency is 200π rad/s.

c) The maximum displacement occurs at the crest of the wave. Since the amplitude is given as 0.05 m, the maximum displacement is also 0.05 m.

The maximum speed of a point on the string occurs at the equilibrium position, where the displacement is zero. At this point, the point is moving with maximum speed. Therefore, the maximum speed is equal to the wave velocity:

Maximum speed = 5 m/s

Hence, the maximum displacement of a point on the string is 0.05 m, and the maximum speed is 5 m/s.