Question

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.25 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.540 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.51 m and t = 0.150 s .

Answer 1

The transverse displacement is
`y(1.51 , 0.150) = 0.055 m`

Step-by-step explanation:

From the question we are told that

The generally equation for the mechanical wave is

`y(x,t) = Acos (kx -wt)`

The speed of the transverse wave is
`v = 8.25 \ m/s`

The amplitude of the transverse wave is
`A = 5.50 *10^(-2) m`

The wavelength of the transverse wave is
`\lambda = 0540 m`

At t= 0.150s , x = 1.51 m

The angular frequency of the wave is mathematically represented as

`w = vk`

Substituting values

`w = 8.25 * 11.64`

`w = 96.03 \ rad/s`

The propagation constant k is mathematically represented as

`k = (2 \pi)/(\lambda)`

Substituting values

`k = (2 * 3.142)/(0. 540)`

`k =11.64 m^(-1)`

Substituting values into the equation for mechanical waves

`y(1.51 , 0.150) = (5.50*10^(-2) ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))`

`y(1.51 , 0.150) = 0.055 m`