Question

Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.

A) 4.55 m.
B) 1.05 m.
C) 3.54 m.
D) 2.25 m.

Answer 1

Step-by-step explanation:

Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.

Amplitude (A) of the simple harmonic wave = 6.44 m

wave number (k) of the given wave = 2.34 m-1

Angular frequency (ω) of the given wave = 2.88 rad/s

Displacement x = 1.21 m and time t = 0.71 s

Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by

y = Asin(kx - ωt)

= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]

Y=6.44sin(0.7866 rad)

0.7866rad*(180 degrees/pi rad) =45.1

Y=6.44sin(45.1)

Y=4.55m

Answer 2

Letter A.
`y=4.55 m`

Step-by-step explanation:

Let's use the wave equation:

`y=Asin(kx-\omega t)`

• A is the amplitude (A=6.44 m)
• t is the time (t=0.71 s)
• k is the wave number (k=2.34 1/m)
• ω is the angular frequency (ω=2.88 rad/s)
• x is the propagation of the x direction (x=1.21 m)

Therefore the displacement y will be:

`y=6.44*sin(2.34*1.21-2.88*0.71)`

`y=4.55 m`

The answer is letter A.

I hope it helps you!