Question:

A transverse wave is traveling along a rope with a velocity of 8 m/s. The amplitude of the wave is 0.5 m and the frequency is 10 Hz.

a) Calculate the wavelength of the wave.

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

c) What is the maximum speed of a point on the rope as the wave passes through?

d) Calculate the maximum acceleration experienced by a particle of the rope.

e) If the rope is under a tension of 4 N, what is the power being transmitted by the wave?

Answer:

a) The wavelength of a wave can be determined using the formula:

wavelength = (velocity) / (frequency)

Given: Velocity (v) = 8 m/s Frequency (f) = 10 Hz

Substituting the given values into the formula:

wavelength = 8 m/s / 10 Hz

wavelength = 0.8 m

Therefore, the wavelength of the wave is 0.8 m.

b) Period (T) represents the time taken for one complete cycle of the wave, while angular frequency (ω) represents the rate at which the wave oscillates in radians per second.

The relationship between period and frequency is: T = 1/f

Given: Frequency (f) = 10 Hz

Substituting the given value into the formula:

T = 1 / (10 Hz)

T = 0.1 s

Therefore, the period of the wave is 0.1 s.

To calculate the angular frequency (ω), we can use the formula:

ω = 2πf

Given: Frequency (f) = 10 Hz

Substituting the given value into the formula:

ω = 2π(10 Hz)

ω ≈ 62.83 rad/s (approximated to two decimal places)

Therefore, the angular frequency of the wave is approximately 62.83 rad/s.

c) The maximum speed of a point on the rope as the wave passes through can be calculated using the formula:

v_max = ω * A

Given: Angular frequency (ω) ≈ 62.83 rad/s Amplitude (A) = 0.5 m

Substituting the given values into the formula:

v_max = 62.83 rad/s * 0.5 m

v_max ≈ 31.42 m/s (approximated to two decimal places)

Therefore, the maximum speed of a point on the rope is approximately 31.42 m/s.

d) The maximum acceleration experienced by a particle of the rope can be calculated using the formula:

a_max = ω^2 * A

Given: Angular frequency (ω) ≈ 62.83 rad/s Amplitude (A) = 0.5 m

Substituting the given values into the formula:

a_max = (62.83 rad/s)^2 * 0.5 m

a_max ≈ 1982.74 m/s^2 (approximated to two decimal places)

Therefore, the maximum acceleration experienced by a particle of the rope is approximately 1982.74 m/s^2.

e) The power being transmitted by the wave can be calculated using the formula:

Power = (1/2) * ρ * A^2 * ω^2 * v

Given: Density of the rope (ρ) = Unknown (Not provided in the question) Amplitude (A) = 0.5 m Angular frequency (ω) ≈ 62.83 rad/s Velocity (v) = 8 m/s

Due to the lack of information about the rope's density, we cannot calculate the exact power being transmitted by the wave without this value.

Therefore, the power being transmitted by the wave cannot be determined with the given information.