Question

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time t = 0, the left end of the string has its maximum upward displacement.


(a) What are the frequency and angular frequency of the wave?

(b) What is the wave number of the wave?

Answer 1

Answer:

(a)

$f=20Hz\\And \\w=125.7s^{-1}$

(b)

$K=3.490m^{-1}$

Explanation:

We know that the speed of any periodic wave is given by:

v=f×λ

The wave number k is given by:

K=2π/λ

Given data

Amplitude A=2.50mm

Wavelength λ=1.80m

Speed v=36 m/s

For Part (a)

For the wave frequency we plug our values for v and λ.So we get:

v=f×λ

$36.0m/s=(1.80m)f\\f=\frac{36.0m/s}{1.80m}\\ f=20Hz$

And the angular speed we plug our value for f so we get:

$w=2\pi f\\w=2\pi (20)\\w=125.7s^{-1}$

For Part (b)

For wave number we plug the value for λ.So we get

K=2π/λ

$K=\frac{2\pi }{1.80m}\\ K=3.490m^{-1}$

Answer 2

The answers for this question are

a.) frequency of the wave is 20 Hz.

b.) angular frequency of the wave is $125.71428\ s^{-1}$

c.) wave number of the wave is $k=3.4920\ m^{-1}$.

Given to us:

Amplitude A= 2.50 mm,

Wavelength λ= 1.80 m,

Velocity V= 36.0 m/sec,

a.) To find out frequency and angular frequency of the wave,we know

$frequency=\dfrac{Velocity}{Wavelength}$

$f=\dfrac{V}{\lambda}$

Putting the numerical value,

$f=\dfrac{36}{1.80}\\\\f= 20\ Hz$

Therefore, the frequency of this wave is 20 Hz.

Also, for angular frequency

$f=2\pi \omega$

Putting the numerical value,

$\omega=2\pi f \\\\\omega=2\times\pi \times 20\\\\\omega= 125.71428\ s^{-1}$

Therefore, the angular frequency of this wave is $125.71428\ s^{-1}$.

b.) To find out the wave number of the wave (k),

The wave number for an EM field is equal to 2π divided by the wavelength(λ) in meters.

$k=\dfrac{2\pi}{\lambda}$

Putting the numerical value,

$k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{1.80}\\\\k=3.4920\ m^{-1}$

Therefore, the wave number of the wave is $k=3.4920\ m^{-1}$.

Hence, for this wave

a.) frequency of the wave is 20 Hz.

b.) angular frequency of the wave is $125.71428\ s^{-1}$

c.) wave number of the wave is $k=3.4920\ m^{-1}$.

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