A teacher sets up a demonstration to show the relationship between circular motion and simple harmonic motion (SHM) - AQA - A-Level Physics - Question 5 - 2022 - Paper 1
Question 5
A teacher sets up a demonstration to show the relationship between circular motion and simple harmonic motion (SHM).
She places a block on a turntable at a point 0.2... show full transcript
Worked Solution & Example Answer:A teacher sets up a demonstration to show the relationship between circular motion and simple harmonic motion (SHM) - AQA - A-Level Physics - Question 5 - 2022 - Paper 1
Step 1
Calculate the time taken for the turntable to complete one revolution.
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Answer
To find the time taken for one complete revolution, we use the formula for angular speed:
Given that one complete revolution corresponds to an angle of 2π radians:
1.8=t2π
Solving for t gives:
t=1.82π≈3.49 s
Step 2
Draw an arrow on Figure 10 to show the direction of the resultant force on the block.
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Answer
An arrow should be drawn pointing towards the center of the turntable, indicating the direction of the resultant centripetal force acting on the block.
Step 3
Calculate the magnitude of the resultant force on the block.
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Answer
To find the magnitude of the resultant force, we can use the formula for centripetal force:
F=mac
where ac=rv2 and v=rω. Thus:
ac=0.25(0.25×1.8)2=0.36m/s2
Substituting:
F=0.12×0.36=0.0432N
Step 4
Describe, with reference to one of Newton's laws of motion, the evidence that a resultant force is acting on the block.
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Answer
According to Newton's first law of motion, an object remains at rest or in uniform motion unless acted upon by a resultant force. Since the block is moving in a circular path, this indicates a changing velocity, which means there must be a resultant force acting on it towards the center of the circular path.
Step 5
Calculate the length of the simple pendulum.
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Answer
The period of the pendulum must match the period of the block's circular motion. The time for one revolution is 2.5 s, so the period T=2.5s.
Using the formula for the period of a pendulum:
T=2πgL
Rearranging to find L:
L=4π2gT2
Given g≈9.81m/s2:
L=4π29.81×(2.5)2≈0.25m
Step 6
Suggest the effect this has on the amplitude relationship and the phase relationship between the moving shadows.
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Answer
Air resistance introduces a damping effect on the pendulum's motion, potentially decreasing its amplitude over time. This could cause the shadows to have different amplitudes, disrupting their original in-phase relationship, and leading to a shift in their phase alignment as the pendulum slows down relative to the constant motion of the block.
