A wave pulse travels down a slinky. The mass of the slinky is m = 0.9 kg and is
ID: 1506493 • Letter: A
Question
A wave pulse travels down a slinky. The mass of the slinky is m = 0.9 kg and is initially stretched to a length L = 7.5 m. The wave pulse has an amplitude of A = 0.21 m and takes t = 0.434 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.49 Hz.
1)
What is the speed of the wave pulse?
m/s
2)
What is the tension in the slinky?
N
3)
What is the average speed of a piece of the slinky as a complete wave pulse passes?
m/s
4)
What is the wavelength of the wave pulse?
m
5)
Now the slinky is stretched to twice its length (but the total mass does not change).
What is the new tension in the slinky? (assume the slinky acts as a spring that obeys Hooke’s Law)
N
6)
What is the new mass density of the slinky?
kg/m
7)
What is the new time it takes for a wave pulse to travel down the slinky?
s
8)
If the new wave pulse has the same frequency, what is the new wavelength?
m
9)
What does the energy of the wave pulse depend on?
the frequency
the amplitude
both the frequency and the amplitude
Explanation / Answer
L = 7.5m
m = 0.9kg
A = 0.21 m
f = 0.49 Hz
t = 0.434 sec
1) v = distance/time = 7.5 / 0.434 = 17.28 m/s
2) T = m*v^2 / L
T = (0.9 x 17.28^2) / 7.5 = 35.83 N
3) average speed = 4*amplitude*frequency
= 4 x 0.21 x 0.49
= 0.4116 m/s
4) wavelength = velocity*period (while period is 1/0.49Hz)
= 17.28*(1/0.49)
= 362.65 m
5) double the distance new velocity = 2L / 0.434 = 34.56 m/s
Tension = (mass * 34.56^2) / 2L = 71.67 N
6) m / 2L = 0.06 kg/m^3
7) t = distance over velocity = 2L / new velocity (34.56m/s) = 0.4340 sec
8) wave length = period * new velocity = (1/0.49Hz) * (34.56m/s) = 70.53 m
9) both frequency and amplitude