a) The time taken for the wave to travel a distance of 8.40 m is 0.24 s
b) The time taken for a point on the string to travel a distance of 8.40 m is 6.46 s
c) The time in part a) does not change; the time in part b) is halved (3.23 s)
Explanation:
a)
In order to solve this part, we have to find the speed of the wave, which is given by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the wave in this problem,
[tex]f = 62.0 Hz\\\lambda =0.560 m[/tex]
So its speed is
[tex]v=(62.0)(0.560)=34.7 m/s[/tex]
Now we want to know how long does it take for the wave to travel a distance of
d = 8.40 m
Since the wave has a constant velocity, we can use the equation for uniform motion:
[tex]t=\frac{d}{v}=\frac{8.40}{34.7}=0.24 s[/tex]
b)
In this case, we want to know how long does it take for a point on the string to travel a distance of
d = 8.40 m
A point on the string does not travel along the string, but instead oscillates up and down. The amplitude of the wave corresponds to the maximum displacement of a point on the string from the equilibrium position, and for this wave it is
A = 5.20 mm = 0.0052 m
The period of the wave is the time taken for a point on the string to complete one oscillation. It can be calculated as the reciprocal of the frequency:
[tex]T=\frac{1}{f}=\frac{1}{62.0}=0.016 s[/tex]
During one oscillation (so, in a time corresponding to one period), the distance covered by a point on the string is 4 times the amplitude:
[tex]d=4A=4(0.0052)=0.0208 m[/tex]
Since d is the distance covered by a point on the string in a time T, then we can find the time t needed to cover a distance of
d' = 8.40 m
By setting up the following proportion:
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0208}=6.46 s[/tex]
c)
As we can see from the equation used in part a), the speed of the wave depends only on its frequency and wavelength, not on the amplitude: therefore, if the amplitude is changed, the speed of the wave is not affected, therefore the time calculated in part a) remains the same.
On the contrary, the answer to part b) is affected by the amplitude. In fact, if the amplitude is doubled:
A' = 2A = 2(0.0052) = 0.0104 m
Then the distance covered during one period is
[tex]d=4A' = 4(0.0104)=0.0416 m[/tex]
The time period does not change, so it is still
T = 0.016 s
Therefore, the time needed to cover a distance of
d' = 8.40 m
this time is
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0416}=3.23 s[/tex]
So, the time taken has halved.
Learn more about waves:
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