A wave pulse travels down a slinky. The mass of the slinky is m = 0.92 kg and is
ID: 1474456 • Letter: A
Question
A wave pulse travels down a slinky. The mass of the slinky is m = 0.92 kg and is initially stretched to a length L = 7.5 m. The wave pulse has an amplitude of A = 0.27 m and takes t = 0.474 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.47 Hz.
1. 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)
2. What is the new mass density of the slinky?
3. What is the new time it takes for a wave pulse to travel down the slinky?
4. If the new wave pulse has the same frequency, what is the new wavelength?
#) What does the energy of the wave pulse depend on?
a)the frequency
b)the amplitude
c)both the frequency and the amplitude
Explanation / Answer
We have two equations in the ondulatory movement
. y= A Sin (k x – w t) v2 = T/ p density symbol: p
With the wave equation v = La f wavelength symbol: La
The speed with which the wave travels is constant
First part
The object like a spring (Hooke's law)
T = -K x X = X0 T= -K (2X0)= 2 T0
If we use the oscillating movement (perpendicular to de wave move)
. w= 2 f
. w2= K/m K= w2 m = 0.92 (2 0.47)2 = 8 N m
T0 = K L = 8 7.5= 60 N
T= 2 60 = 120 N
Second part
We are looking for the new density
. v2 = T/r p = T/ v2
. v= La f
We are looking for wavelength
. k= 2 /l w= 2 f T=1/f
. y= A Sin (k x – w t)= A sin ( 2 (x/La -t/T) y= 0.27 Sin (2 ( x/La - t/2.12 )
. t= T= 0.47 s x=L
For the point initial the amplitude y=0 we have Sin (kx-wt)= 0
. 2 (L/l -T/2.12) = 2 L/l - T/T = 1 x/l=2 l= 2/L=2/7.5=0.266 m
Introducing
. v = 0.266 0.47 = 0.125 m/s
p = T/ v2 = 2 T0 / v2 = 2 60/ 0.1252 = 7680 Kg/m3
Third part
As the speed is constant
. v=d/t t= d/v= 2L0/ v t= 2 7.5/ 0.125 = 120