WAVES&OPTICS: Given a snapshot graph of a standing wave that has an amplitude of 1 at t=0ms, what will the amplitudes be for the subsequent snapshot graphs at t=1,3,5,7ms?

Question

"A snapshot at t=0ms of the m=2 mode of a fixed-fixed string is show below. The period of this standing wave is t=8ms. Draw snapshot graphs of the string every 1ms from t=1 ms to t=8 ms. Think carefully about the amplitude at each instant."

A graph is given with an inverted sine wave (so starting at 0) with an amplitude A (which can be replaced with 1 for this problem).

I've figured out already that at t=2 ms and 6ms, the wave is at equilibrium and therefore a horizontal line, at t=4 ms the wave is the inverted version of the original wave (max. positive displacement), and that at t=8 ms the wave returns to its starting position because of the wave's period.

Also at t=0 ms (the given graph) the wave is at max. negative displacement.

Answer 1

To determine the amplitudes of the standing wave at different times, we need to understand the behavior of the standing wave over time. The standing wave can be described by the equation:y(x,t) = Asin⁡(kx) cos⁡(ωt)

where: A is the amplitude of the wave, k is the wave number, ω is the angular frequency, x is the position along the string, t is time. Given that the period T of the standing wave is 8 ms, the angular frequency ω is:ω = 2π/T = 2π/8 ms = π/4 rad/ms At t=0 ms, the wave is at its maximum negative displacement, which means:y(x,0) = A sin⁡(kx) cos⁡(0) = A sin(kx) Since the wave is at maximum negative displacement, we can infer that sin(kx)=−1 at the points of maximum displacement.

Now, let's determine the amplitude at different times: At t=1 ms: cos⁡((π/4)×1)=cos⁡(π/4)= Sqrt(2)/2​​

The amplitude at t=1 ms is: y(x,1)= A sin⁡(kx) cos⁡(π/4) = A sin⁡(kx) x Sqrt(2)/2​​

Since sin(kx)=−1 at maximum displacement: y(x,1)=−A×Sqrt(2)/2​​=−ASqrt(2)/2​

At t=2 ms: cos⁡(π/4×2)=cos⁡(π/2)=0 The amplitude at t=2 ms is: y(x,2)= A sin⁡(kx)cos⁡(π/2)=Asin⁡(kx)×0=0

The wave is at equilibrium. At t=3 ms: cos⁡(π/4×3)=cos⁡(3π/4)=−Sqrt(2)/2​​

The amplitude at t=3 ms is: y(x,3)=Asin⁡(kx)cos⁡(3π/4) = Asin⁡(kx)×(−Sqrt(2)/2​​)

Since sin(kx)=−1 at maximum displacement: y(x,3)=−A×(−Sqrt(2)/2​​)= AxSqrt(2)/2​​

At t=4 ms: cos⁡(π/4×4)=cos⁡(π)=−1cos(4/π​×4)=cos(π)=−1 The amplitude at t=4 ms is: y(x,4)=Asin⁡(kx)cos⁡(π)=Asin⁡(kx)×(−1) Since sin(kx)=−1 at maximum displacement: y(x,4)=−A×(−1)=A

The wave is at maximum positive displacement.

At t=5 ms: cos⁡((π/4)×5)=cos⁡(5π/4)=−Sqrt(2)/2​​

The amplitude at t=5 ms is: y(x,5)=Asin⁡(kx)cos⁡(5π/4)=Asin⁡(kx)×(−Sqrt(2)/2​​) Since sin⁡(kx)=−1 at maximum displacement: y(x,5)=−A×(−Sqrt(2)/2​​)=AxSqrt(2)/2​​

At t=6 ms: cos⁡(π/4×6)=cos⁡(3π/2)=0

The amplitude at t=6 ms is: y(x,6)=Asin⁡(kx)cos⁡(3π/2)=Asin⁡(kx)×0=0

The wave is at equilibrium. At t=7 ms: cos⁡((π/4)×7)=cos⁡(7π/4)=Sqrt(2)/2

The amplitude at t=7 ms is: y(x,7)=Asin⁡(kx)cos⁡(7π/4)=Asin⁡(kx)×Sqrt(2)/2

Since sin⁡(kx)=−1 at maximum displacement:y(x,7)=−A×Sqrt(2)/2=−AxSqrt(2)/2

At t=8 ms: cos⁡((π/4)×8)=cos⁡(2π)=1 The amplitude at t=8 ms is: y(x,8)=A sin⁡(kx) cos⁡(2π)= Asin⁡(kx)×1 Since sin⁡(kx)=−1 at maximum displacement: y(x,8)=−Ay(x,8)=−A The wave returns to its starting position.