Suppose that the range of output frequencies is from 89.2 Hz to 13.9 kHz for a pipe organ. Take 343 m/s for the speed of sound.
1 answer:
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(a) 0.96 m/s
The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:
The frequency of the waves is the reciprocal of the period:
The wavelength instead is just the distance between two consecutive crests, so
And the wave speed is given by:
(b) 0.265 m
The total distance between the highest point of the wave and its lowest point is
d = 0.53 m
The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is
(c) Amplitude: 0.15 m, wave speed: same as before
In this case, the amplitude of the wave would be lower. In fact,
d = 0.30 m
So the amplitude would be
Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.
Answer:
Explanation:
Angular acceleration is defined by
Angular velocity is related to the period by
Putting all together:
Taking our initial (i) point now and our final (f) point one year later, we would have:
So for our values we have:
Where the minus sign indicates it is decelerating.