Question:

A wave travels along a stretched string with a speed of 4 m/s. The amplitude of the wave is 0.2 m and the frequency is 50 Hz. Determine the wavelength, period, and wave speed of this wave.

Answer:

Given: Speed of the wave (v) = 4 m/s Amplitude (A) = 0.2 m Frequency (f) = 50 Hz

We are to find: Wavelength (λ), Period (T), and Wave speed (v)

1. Wavelength (λ): The formula relating wavelength, frequency, and wave speed is given by: v = f * λ

Rearranging the formula, we can find the wavelength (λ): λ = v / f

Substituting the given values into the formula, we have: λ = 4 m/s / 50 Hz λ = 0.08 m

Therefore, the wavelength of the wave is 0.08 m.

2. Period (T): The period (T) is the time taken for one complete wave to pass a fixed point. It is inversely proportional to the frequency (f) and is given by the formula: T = 1 / f

Substituting the given value of frequency (f) into the formula, we have: T = 1 / 50 Hz T = 0.02 s

Therefore, the period of the wave is 0.02 s.

3. Wave speed (v): The wave speed (v) is determined by the medium through which the wave is traveling. It can be found by multiplying the wavelength (λ) by the frequency (f): v = λ * f

Substituting the given values into the formula, we have: v = 0.08 m * 50 Hz v = 4 m/s

Therefore, the wave speed of the wave is 4 m/s.

In conclusion, the wavelength of the wave is 0.08 m, the period is 0.02 s, and the wave speed is 4 m/s.