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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

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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 can use the formula:

$T = \frac{2\pi}{\omega}$

where $T$ is the time period and $\omega = 1.8$ rad s^-1 is the angular speed.

Calculating:

$T = \frac{2\pi}{1.8} \approx 3.49 \text{ s}$

Step 2

Draw an arrow on Figure 10 to show the direction of the resultant force on the block.

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Answer

The 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

The resultant force can be calculated using the formula for centripetal force:

$F = m a_c$

where $m = 0.12$ kg is the mass of the block and $a_c = \frac{v^2}{r}$ is the centripetal acceleration. To find $v$ , we can use $v = r \omega = 0.25 \times 1.8 = 0.45$ m/s. Thus:

$a_c = \frac{(0.45)^2}{0.25} = 0.81 \text{ m/s}^2$

Now calculating the force:

$F = 0.12 \times 0.81 \approx 0.097 \text{ N}$

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 at rest will stay at rest and an object in motion will stay in motion unless acted upon by a resultant force. In this case, the block moves in a circular path, indicating a constant change in direction, which can only occur if a resultant force is acting upon it to keep it in circular motion.

Step 5

Calculate the length of the simple pendulum.

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Answer

For a pendulum to complete one full oscillation in the same time period as the rotating block (2.5 s), we use the formula for the period of a simple pendulum:

$T = 2\pi \sqrt{\frac{L}{g}}$

Rearranging this gives:

$L = \frac{g T^2}{4\pi^2}$

Substituting $g = 9.81$ m/s² and $T = 2.5$ s:

$L = \frac{9.81 \times (2.5)^2}{4\pi^2} \approx 0.84 ext{ m}$

Step 6

Suggest the effect this has on the amplitude relationship and the phase relationship between the moving shadows.

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Answer

The air resistance will dampen the motion of the pendulum, resulting in reduced amplitude over time. This change may lead to a phase difference between the shadows since the block continues its motion uniformly, while the pendulum’s motion becomes slower and less pronounced.

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Questions answered

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

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

Calculate the time taken for the turntable to complete one revolution.

Draw an arrow on Figure 10 to show the direction of the resultant force on the block.

Calculate the magnitude of the resultant force on the block.

Describe, with reference to one of Newton's laws of motion, the evidence that a resultant force is acting on the block.

Calculate the length of the simple pendulum.

Suggest the effect this has on the amplitude relationship and the phase relationship between the moving shadows.

Join the A-Level students using SimpleStudy...

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