A wave pulse travels down a slinky. The mass of the slinky is m = 0.89 kg and is initially stretched to a length L = 6.2 m. The wave pulse has an amplitude of A = 0.27 m and takes t = 0.408 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.44 Hz.1) What is the speed of the wave pulse? 17.035 m/s
2) What is the tension in the slinky? 33.92 N
3) What is the average speed of a piece for the slinky as a complete wave pulse passes? .4704 m/s
4) What is the wavelength of the wave pulse? 34.765 m

Respuesta :

1) The wave speed is 15.2 m/s

2) The tension in the slinky is 33.2 N

3) The average speed of a piece of the slinky during one pulse is 0.475 m/s

4) The wavelength is 34.5 m

Explanation:

1)

The motion of a wave pulse along the slinky is a uniform motion, therefore its speed is given by the equation for uniform motion:

[tex]v=\frac{L}{t}[/tex]

where

L is the length covered

t is the time elapsed

For the wave in this problem, we have:

L = 6.2 m is the length of the slinky

t = 0.408 s is the time taken for a pulse to travel across the length os the slinky

Substituting, we find the wave speed

[tex]v=\frac{6.2}{0.408}=15.2 m/s[/tex]

2)

The speed of a wave on a slinky can be found with the same expression for the wave speed along a string:

[tex]v=\sqrt{\frac{T}{m/L}}[/tex]

where

T is the tension in the slinky

m is the mass of the slinky

v is the wave speed

L is the length

In this problem, we have:

m = 0.89 kg is the mass of the slinky

L = 6.2 m is the length

Therefore, we can re-arrange the equation to find the tension in the slinky, T:

[tex]T=v^2 (\frac{m}{L})=(15.2)^2 (\frac{0.89}{6.2})=33.2 N[/tex]

3)

The average speed of a piece of the slinky as a complete wave pulse passes is the total displacement done by a piece of slinky during one period, which is 4 times the amplitude, divided by the time taken for one complete oscillation, the period:

[tex]v_{avg} = \frac{4A}{T}[/tex]

where

A is the amplitude

T is the period

Here we have:

A = 0.27 m is the amplitude of the wave

The period is the reciprocal of the frequency, therefore

[tex]T=\frac{1}{f}[/tex]

where f = 0.44 Hz is the frequency of this wave. Substituting and solving, we find

[tex]v_{avg} = \frac{4A}{1/f}=4Af=4(0.27)(0.44)=0.475 m/s[/tex]

4)

The wavelength of the wave pulse can be found by using the wave equation:

[tex]v=f\lambda[/tex]

where

v is the wave speed

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the pulse in this problem, we have

v = 15.2 m/s

f = 0.44 Hz

Substituting, we find the wavelength:

[tex]\lambda=\frac{v}{f}=\frac{15.2}{0.44}=34.5 m[/tex]

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