Question

A 3 m slinky rests on a table with a student holding each end. The students laterally shake the ends of the slinky to generate transverse waves. The student on the left shakes the slinky at 4 Hz with a 5 cm amplitude and the student on the right shakes it at 3 Hz with a 7 cm amplitude. What is the maximum deflection of the string, from its resting position, between the students in cm?

Answer 1

To find the maximum deflection of the string between the students, we need to calculate the superposition of the two waves generated by each student.

The equation for a transverse wave is given by:

y(x, t) = A * sin(kx - ωt + ϕ)

where:
- y is the displacement of the string at position x and time t
- A is the amplitude of the wave
- k is the wave number
- ω is the angular frequency
- ϕ is the phase angle

For the student on the left shaking the slinky at 4 Hz with a 5 cm amplitude, the wave equation is:

y1(x, t) = 5 * sin(2π * 4 * t - 2π * x/3 + ϕ1)

For the student on the right shaking the slinky at 3 Hz with a 7 cm amplitude, the wave equation is:

y2(x, t) = 7 * sin(2π * 3 * t - 2π * x/3 + ϕ2)

The maximum deflection between the two students occurs when the two waves are in phase. This happens when the total phase difference between the two waves is an integer multiple of 2π. So we need to find the position x at which the two waves will be in phase.

For the two waves to be in phase:
2π * 4 * t - 2π * x/3 + ϕ1 = 2π * 3 * t - 2π * x/3 + ϕ2
2π * (4 - 3) * t = ϕ2 - ϕ1
t = (ϕ2 - ϕ1)/(2π)

For simplicity, let's assume that the phase difference ϕ2 - ϕ1 is zero (i.e., the two waves are initially in phase).

Substitute t = 0 into the wave equations to find the max deflection from the resting position:

y_max(x) = y1(x, 0) + y2(x, 0)
y_max(x) = 5 * sin(ϕ1) + 7 * sin(ϕ2)

Given the amplitudes are 5 cm and 7 cm, and assuming that the initial phase angles are 0, we have:

y_max(x) = 5 + 7 = 12 cm

Thus, the maximum deflection of the string, from its resting position, between the students is 12 cm.