Question

A wave is traveling on a string with a velocity of 15 m/s. The frequency of the wave is 60 Hz and its wavelength is 0.25 m. Determine the amplitude, period, and angular frequency of this wave.

Answer

The velocity of a wave is given by the equation:

v = λf

Where:

• v is the velocity of the wave,
• λ is the wavelength, and
• f is the frequency.

We are given v = 15 m/s and λ = 0.25 m. We can rearrange the equation to solve for f:

f = v / λ

Substituting the given values:

f = 15 m/s / 0.25 m = 60 Hz

So the frequency of the wave is 60 Hz.

The period of a wave is the reciprocal of its frequency:

T = 1 / f

Substituting the frequency we just found:

T = 1 / 60 Hz = 0.0167 s

So the period of the wave is approximately 0.0167 s.

The angular frequency (ω) of a wave is related to its frequency by the equation:

ω = 2πf

Substituting the given frequency:

ω = 2π(60 Hz) = 120π rad/s

Now, we can use the formula for the velocity of a wave to find its amplitude (A):

v = ωA

Rearranging the equation to solve for A:

A = v / ω

Substituting the given values:

A = 15 m/s / 120π rad/s ≈ 0.039 m

So, the amplitude of the wave is approximately 0.039 m.

Therefore, the amplitude is 0.039 m, the period is approximately 0.0167 s, and the angular frequency is 120π rad/s.