A transverse periodic wave is established on a string. The wave is described by the expression y = 0.005 sin(20.0x + 2ft) where y is in meters when x and t are in meters and seconds, respectively. If the wave travels with a speed of 20.0 m/s, what is its frequency, f?

f =63.64Hz

Explanation:

The equation of the transverse periodic wave

y = 0.005 sin(20.0x + 2ft)

From the equation we can deduce that

Amplitude A = 0.005

Wave number K = 20

Angular frequency w = 2πf

We are asked to find the frequency (f)

the formular for the wave number is given as K = 2π / lambda

Lambda = 2π/ K

Lamda is the wavelength

= 2 × 3.142 /20

= 0.3142m

The velocity of the wave is expressed as v = frequency × lamba (wavelength)

Frequency = velocity / wavelength

f = 20/ 0.3142

f = 63.65Hz

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explanation:

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