Vinyes Problem 20.39 Resources previous | 18 of 30n Problem 20.39 Part A You hav
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Vinyes Problem 20.39 Resources previous | 18 of 30n Problem 20.39 Part A You have two vibrating objects in an infinitely large pool. The distance between them is 54 m. Their frequency of vibration is 2.0 Hz and the wave speed is 3.0 m/s. The vibrations are sinusoidal. Find a location between them where the water does not vibrate. Assume n 0. Express your answer to two significant figures and include the appropriate units. Value Units Submit My Answers Give Up Part B Find another location between them where the water vibrates with the largest amplitude. Assume Express your answer to two significant figures and include the appropriate units. HA Fntinode Vlue Units Submit My Answers Give Up Provide Feedback ContinueExplanation / Answer
If we think of a string tied between the two sources, in a resonance at the given frequency and wavelength, the in-phase positions correspond to the nodes of the string, and the out-of-phase positions correspond to the antinodes
The expression for the value of phase along some point of a sinusoidal wave is:
= kx-t
where
k = (angular) wavenumber = 2/, where =wavelength
= angular frequency = 2f, where f = (linear) frequency in Hz
So, at a point a distance d from one source, along a line between the two sources, the wave from each source has a phase
1 = kd - t ( "Source 1" )
2 = k(6-d)-t
two sine waves will be in phase if the magnitude of the phase difference between them at a given point is equal to either 0 or an integer multiple of 2.
|2-1|= |(kd-t)-(k(6-d)-t)|= |k(d-(6-d))| =| k(2d-6)| = |((2)/)(2d-6)| = 2m
The factors of 2 on each side cancel out, leaving:
|(2d-6)/| = m, with m = 0,1,2,3...
The value for the wavelength can be calculated from the given quantities, using:
v=f --> 4 m/s = (2 Hz)() --> = 2 m
This means, for the two waves being in phase between the sources, we have:
|(2d-6)/2| = m (m= 1,2,3...) --> |d-3| = m
we need to start at m = 1, and continue solving for d for each allowed value of m, until d exceeds 6 m. This means the allowed values for d, for in phase, are:
d = 1 m, 2 m, 3 m, 4 m, 5 m
For second part,
they are not in phase, they are "out of phase," but, if they mean 180o or radians out of phase, we just have to change the condition of the magnitude of the phase difference from being 0 or an integer multiple of 2 to being an odd-integer multiple of . i.e
|d-3| = (m+1/2) with m=0,1,2,3...
Starting at m = 0, we cycle through the allowed values of m and note the solutions for d that do not exceed 6 m. For this case, then, we have allowed solutions for out-of-phase d of:
d = 0.5, 1.5, 2.5, 3.5, 4.5, 5.5 m
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