A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m + t/(0.20 s))], where x is in m and t in s.
In what direction is this wave traveling?
What are the wave speed, frequency, and wavelength?
At t = 0.50 , what is the displacement of the string at x = 0.20 m?

Respuesta :

Corrected and Formatted Question:

A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.

(a) In what direction is this wave traveling?

(b) What are the wave speed, frequency, and wavelength?

(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?

Answer:

The wave is travelling in the negative x direction

The wave speed = 12.0m/s

The frequency = 5Hz

The wavelength = 2.4m

The displacement at t = 0.50s and x = 0.20m is -0.029m

Explanation:

The general wave equation is given by;

y(x, t) = y cos (2[tex]\pi[/tex](x/λ) - 2[tex]\pi[/tex]ft)    --------------------------------(i)

Where;

y(x, t) is the displacement of the wave at position x and a given time t

y = amplitude of the wave

f = frequency of the wave

λ = wavelength of the wave

Given;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]   ------------------(ii)

Which can be re-written as;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))]  -------------(iii)

Comparing equations (i) and (iii) we have that;

=> 2π(x/(2.4 m) = 2π(x/λ)

=> λ = 2.4m

Therefore the wavelength of the wave is 2.4m

Also, still comparing the two equations;

=> 2π(t/(0.20 s) = 2πft

=> f = 1 / 0.20

=> f = 5Hz

Therefore the frequency of the wave is 5Hz

To get the wave speed (v), it is given by;

v = f x λ

Where f = 5Hz and λ = 2.4m

=> v = 5 x 2.4

=> v = 12.0m/s

Therefore, the speed of the wave is 12.0m/s

At t = 0.50s and x = 0.20m;

The displacement, y(x,t) of the string wave is given by

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]

Convert the amplitude of 3.0cm to m

=> 3.0cm = 0.03m

Substitute this back into the equation

=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]

Substitute the values of t and x into the equation above;

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]

Carefully solve the equation

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]

=> y(x, t) = (0.03m) × cos[0.08π + 5π]

=> y(x, t) = (0.03m) × cos[5.08π]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × -0.9684

=> y(x, t) = 0.029m

Therefore the displacement at those points is -0.029m

Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.