Question

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.0 vibrations in 33.0 s. Also, a given maximum travels 421 cm along the rope in 8.0 s. What is the wavelength?

Answer 1

Answer:

0.47 m

Explanation:

$N$ = Number of vibrations = 37

$t$ = total time taken = 33 s

$T$ = time period of each vibration

frequency of vibration is given as

$f = \frac{N}{t} \\f = \frac{37}{33} \\f = 1.12$ Hz

$d$ = distance traveled along the rope = 421 cm = 4.21 m

$t$ = time taken to travel the distance = 8 s

$v$ = speed of the wave

Speed of the wave is given as

$v = \frac{d}{t}\\v = \frac{4.21}{8}\\v = 0.53 ms^{-1}$

$\lambda$ = wavelength of the harmonic wave

wavelength of the harmonic wave is given as

$\lambda = \frac{v}{f} \\\lambda = \frac{0.53}{1.12} \\\lambda = 0.47 m$

Answer 2

Answer:

0.472 m

Explanation:

f = cycles/ second = 33 / 37 = 0.892 Hz

Vmean = Δx / Δt = 4.21m / 8s = 0.526

Vwave = λ x f

⇒ λ = f / Vwave = 0.526 / 0.892 = 0.472 m