Question

A harmonic wave travels in the positive x direction at 12 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation
y = 0.0205 cos(4t)
where y is the displacement in meters and the time t is in seconds
(a) What is the amplitude of the wave, in meters?
(b) What is the frequency of the wave, in hertz?
(c) What is the wavelength of the wave, in meters?

Answer 1

Answer:

a)A= 0.0205  m

b) f = 0.636 Hz

c)λ = 1.57 m

Explanation:

Given that

Velocity v= 12 m/s

y=  0.0205 cos(4t)             ----------1

The standard form of the wave equation given as

y= A cos( ωt - k x)

If point is fix then  equation become

y= A cos ωt            -------------2

By  comparing equation 1 and 2

Amplitude A

A= 0.0205  m

ω = 4 rad/s

We know that

ω = 2 π f

f=Frequency

4 = 2 π f

f = 0.636 Hz

The wavelength

The velocity V= ω k

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

$V=\omega \dfrac{2\pi}{\lambda }$

$12=3\times  \dfrac{2\pi}{\lambda }$

λ = 1.57 m

Answer 2

The amplitude of the wave, in meters, is 0.0205m

The frequency of the wave, in meters is  2/π hertz

The wavelength of the wave, in meters, is 2π/3 meters

The formula for the wave equation is expressed as:

y = Acos(wt + kx)

If a point on the string is fixed, then the equation of the wave will be expressed as:

y = Acos(wt)

A is the amplitude of the wave

w is the angular velocity

t is the time taken

Given the wave equation y = 0.0205 cos(4t)

Comparing with the general equation:

A = 0.0205m

Hence, the amplitude of the wave, in meters is 0.0205m

w = 4

2πf = 4

f = 4/2π

f = 2/π hertz

The frequency of the wave, in meters is  2/π hertz

Similarly,

w = vk

v = w(2π/λ)

12 = 4(2π/λ)

12 = 8π/λ

λ = 8π/12

λ = 2π/3

Hence the wavelength of the wave, in meters is 2π/3 meters

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