Answer:
(a) 0.177 m
(b) 16.491 s
(c) 25 cycles
Explanation:
(a)
Distance between the maximum and the minimum of the wave = 2A ............ Equation 1
Where A = amplitude of the wave.
Given: A = 0.0885 m,
Distance between the maximum and the minimum of the wave = (2×0.0885) m
Distance between the maximum and the minimum of the wave = 0.177 m.
(b)
T = 1/f ...................... Equation 2.
Where T = period, f = frequency.
Given: f = 4.31 Hz
T = 1/4.31
T = 0.23 s.
If 1 cycle pass through the stationary observer for 0.23 s.
Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.
= 16.491 s.
(c)
If 1.21 m contains 1 cycle,
Then, 30.7 m will contain (30.7×1)/1.21
= 25.37 cycles
Approximately 25 cycles.
The list of choices you provided with your question
is utterly devoid of any such examples.
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Answer:
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