1.A pendulum has a period of 0.58s on Earth. Part A What is its period on Mars,

Question

1.A pendulum has a period of 0.58s on Earth.

Part A

What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

Express your answer using two significant figures.

2.A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 3.00s for the boat to travel from its highest point to its lowest, a total distance of 0.600m . The fisherman sees that the wave crests are spaced a horizontal distance of 6.00m apart.

Part A

How fast are the waves traveling?

Express the speed v in meters per second using three significant figures.

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Part B

What is the amplitude A of each wave?

Express your answer in meters using three significant figures.

  m  

3.

3.A person standing a certain distance from an airplane with four equally noisy jet engines is experiencing a sound level of 141dB .

Part A

3.What sound level would this person experience if the captain shut down all but one engine? [Hint: Add intensities, not dB's.]

Express your answer to three significant figures and include the appropriate units.

? =

4.As a bat flies toward a wall at a speed of 6.50m/s ,the bat emits an ultrasonic sound wave with frequency 28.0kHz .

Part A

What frequency does the bat hear in the reflected wave?

Explanation / Answer

1)Time period of a pendulum is inversely proportional to the under root of acceleration due to gravity.
Te/Tm=(gm^1/2)/(ge^1/2)
.58/Tm=(.37)^1/2
Tm is approximately 0.9535 seconds

2)

So, from the data given:
Time for wave to travel from highest point to lowest = 3.0 sec.
This corresponds to half of one wavelength.
Therefore, time for one whole wavelength = 6.0 sec.

The wave crests are spaced 6.0 m. apart.
This corresponds to the distance from one peak to the next peak of the waves.
By definition, this is one whole wavelength.
Hence, one wavelength = 6.0 m.

Since the speed of the wave = distance / time, we have speed = 6.00 m. divided by 6.0 sec.
And that is 1.0 m / sec.; the speed of the wave in a horizontal direction.

As for the amplitude of the wave, the highest point of the wave to its lowest point is 0.600 m (The boat always sits on the wave - I assume it isn't sinking!)
This distance corresponds to the peak - peak vertical displacement of the wave.
Therefore, the amplitude is half of this, i.e. 0.3 m

3)

You need to divide the power level by 4, however you must use the logrithm rules as well as the fact that dB = 10*log base 10 (power) so

10 log( power / 4) = 10 log(power) - 10 log(4)

we already have 10 log(power) = 141 so we figure out that 10 log(4) = 6.02 So the dB power of 1 engine would be 141 - 6 = 135 dB

4)

The wall gets a sound higher than the bat sends out because the bat is flying towards it.
The change in pitch / original frequency = 6.5 m/Sec / speed of sound
Delta F= 28000 * 6.5 /300 or 606.67 Hz.

But the bat hears this frequency emitted from the wall and coming towards it.
So it hears a new delta F = 28606.67 * 6.5 / 300 or 619.81 Hz

So the frequency the bat hears is 28606.67 + 619.81 or 29226.48Hz
or 29.226 KHz

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