A 3m 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?
To find the maximum deflection of the string, we need to find the point of maximum constructive interference between the two waves generated by the students.
The wave generated by the student on the left can be represented by the equation:
y1(x, t) = A1 * sin(k1*x - ω1*t)
where:
A1 = 5 cm is the amplitude of the left wave
k1 = 2π / λ1 is the wave number of the left wave
ω1 = 2π * f1 is the angular frequency of the left wave
The wave generated by the student on the right can be represented by the equation:
y2(x, t) = A2 * sin(k2*x - ω2*t)
where:
A2 = 7 cm is the amplitude of the right wave
k2 = 2π / λ2 is the wave number of the right wave
ω2 = 2π * f2 is the angular frequency of the right wave
Since the two waves are in phase, the point of maximum constructive interference occurs when the two waves are in phase and at their maximum amplitude.
To determine when the two waves are in phase, we need to find the phase difference between the two waves:
Δϕ = (k1*x - ω1*t) - (k2*x - ω2*t)
= (k1 - k2)*x - (ω1 - ω2)*t
The waves will be in phase when the phase difference is zero:
(k1 - k2)*x - (ω1 - ω2)*t = 0
This equation can be rearranged to solve for x:
x = (ω1 - ω2) * t / (k1 - k2)
The maximum deflection can be found by substituting the time at which the two waves are in phase into either of the wave equations.
Since the maximum deflection is the amplitude of the wave, we can find it by substituting x = 0 into one of the wave equations.
Let's substitute the values into the equations:
For the left wave:
A1 * sin(k1 * 0 - ω1 * t) = A1 * sin(- ω1 * t)
For the right wave:
A2 * sin(k2 * 0 - ω2 * t) = A2 * sin(- ω2 * t)
Since sin(-θ) = -sin(θ), the maximum deflection can be represented as:
Dmax = -A1 * sin(ω1 * t) + A2 * sin(ω2 * t)
To find the time at which the two waves are in phase, we need to find the least common multiple (LCM) of the two frequencies.
LCM(4 Hz, 3 Hz) = 12 Hz
So, the waves will be in phase every 1/12 of a second.
Substituting t = 1/12 seconds into the equation for maximum deflection:
Dmax = -A1 * sin(ω1 * (1/12)) + A2 * sin(ω2 * (1/12))
The angular frequencies can be calculated using the given frequencies:
ω1 = 2π * f1 = 2π * 4 Hz = 8π Hz
ω2 = 2π * f2 = 2π * 3 Hz = 6π Hz
Substituting the values:
Dmax = -5 cm * sin(8π * (1/12)) + 7 cm * sin(6π * (1/12))
Simplifying:
Dmax = -5 cm * sin(2π/3) + 7 cm * sin(π/2)
Using the values of sine for these angles:
Dmax = -5 cm * (sqrt(3)/2) + 7 cm * 1
Simplifying:
Dmax = -5sqrt(3)/2 + 7 cm
Therefore, the maximum deflection of the slinky between the students is approximately -5sqrt(3)/2 + 7 cm.