Pendulum Time Warp: A Quiz On The Period Of A Simple Pendulum
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Have you studied the Simple Pendulum in your Physics class? Take this quiz on the time period of a simple pendulum, and test your knowledge. We have got a few questions for your practice where you can see how much you know and how much more you need to learn. Go for it and try for a perfect score. After that, you can share the quiz with your friends and challenge them on this quiz. Just check out who scores better!
Questions and Answers
1.
Hamiz and Haziq did the experiment on Period of the pendulum.
They conclude that "as the length of the string decreases, the period of the pendulum also decreases."
Do you agree with their conclusion on the experiment?
A.
True
B.
False
Correct Answer
A. True
Explanation The conclusion that "as the length of the string decreases, the period of the pendulum also decreases" is correct. This is because the period of a pendulum is directly proportional to the square root of the length of the string. As the length of the string decreases, the period of the pendulum decreases as well. This can be explained by the fact that a shorter string allows the pendulum to swing back and forth more quickly, resulting in a shorter period. Therefore, it is reasonable to agree with their conclusion.
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2.
One oscillation of a swinging pendulum occurs when the bob moves from X and Y and back to X again. Using a stopwatch, which would be the most accurate way to measure the time for one oscillation of the pendulum?
A.
Time 20 oscillations and divide by 20
B.
Time 20 oscillation and multiply by 20
C.
Time one oscillation
D.
Time the motion from X and Y, and double it
Correct Answer
A. Time 20 oscillations and divide by 20
Explanation To measure the time for one oscillation of the pendulum accurately, timing 20 oscillations and then dividing the total time by 20 would be the most accurate method. This approach helps to minimize any errors caused by reaction time or human error in starting and stopping the stopwatch precisely at the beginning and end of each oscillation. By averaging the time of multiple oscillations, the accuracy of the measurement is improved.
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3.
Suppose a simple pendulum consists of a 1.2 m string and a 0.5 kg bob, and its period is 3.75s.
What would be the period of the pendulum if the mass of the bob were doubled?
(don't forget to put the unit)
Explanation The period of a simple pendulum is not affected by the mass of the bob. It is only determined by the length of the string and the acceleration due to gravity. Therefore, even if the mass of the bob is doubled, the period of the pendulum will remain the same, which is 3.75 seconds.
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4.
A pendulum swings backward and forwards, passing through the center, the middle point of the oscillation. The first time the pendulum passes through the middle point, a stopwatch is started. The Twenty-first time the pendulum passes through the middle point, the stopwatch is stopped.
The reading is T.
WHAT IS THE PERIOD OF THE PENDULUM?
A.
T/10
B.
T/20
C.
T/21
D.
T/40
Correct Answer
A. T/10
Explanation The period of a pendulum is the time it takes for one complete oscillation. In this scenario, the stopwatch is started when the pendulum passes through the middle point for the first time and stopped when it passes through the middle point for the twenty-first time. Therefore, the time measured by the stopwatch represents the time for 20 oscillations. To find the period of the pendulum, we divide the time measured by the stopwatch (T) by the number of oscillations (20). Therefore, the correct answer is T/20, which represents the period of the pendulum.
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5.
Suppose that it takes a simple pendulum 1.8 seconds to swing from its leftmost point to its rightmost point. What is the period of the pendulum?
Explanation The period of a pendulum is the time it takes for one complete swing, from the leftmost point to the rightmost point and back to the leftmost point. Since it takes the simple pendulum 1.8 seconds to swing from one side to the other, the period can be calculated by multiplying this time by 2, resulting in 3.6 seconds.
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6.
A simple pendulum of time period TT is taken inside a car moving with a constant velocity of 50m/s on a straight road. What will be the time period of the pendulum inside the moving car, assuming everything to be ideal?
A.
Increases
B.
Decreases
C.
Remains same
D.
Insufficient data
Correct Answer
C. Remains same
Explanation The time period of a simple pendulum is determined by the length of the pendulum and the acceleration due to gravity. Since the car is moving with a constant velocity, there is no change in the acceleration due to gravity. Therefore, the time period of the pendulum will remain the same inside the moving car.
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7.
If the length of a simple pendulum is doubled. What will happen to its period?
A.
Increase by a factor of sqrt(2)
B.
Halve
C.
Double
D.
Decrease by a factor of sqrt(2)
Correct Answer
A. Increase by a factor of sqrt(2)
Explanation When the length of a simple pendulum is doubled, its period will increase by a factor of sqrt(2). The period of a pendulum is directly proportional to the square root of its length. By doubling the length, the square root of 2 is introduced into the equation, resulting in an increase in the period.
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8.
For a simple pendulum, the time period is π seconds. Its velocity is 10 m/s at the instant when it crosses its mean position. What is the amplitude of oscillation?
A.
1 m
B.
5 m
C.
10 m
D.
20 m
Correct Answer
D. 20 m
Explanation The time period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. In this case, the time period is given as π seconds. By rearranging the formula, we can solve for L and find that L = (T/2π)^2 * g. Since T = π seconds and g is a constant, we can conclude that the length of the pendulum is constant. Therefore, the amplitude of oscillation, which is the maximum displacement from the mean position, is equal to the length of the pendulum, which is 20 m.
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9.
A net force of 430 N acts to stretch an extension spring. How far will the spring stretch if its spring constant is 700 N/m?
A.
42.5 m
B.
0.614 m
C.
0.305 m
D.
1.54 m
Correct Answer
B. 0.614 m
Explanation The spring constant represents the stiffness of the spring, indicating how much force is required to stretch or compress it. In this case, the spring constant is given as 700 N/m. The formula to calculate the displacement of the spring is F = kx, where F is the force applied, k is the spring constant, and x is the displacement. Rearranging the formula to solve for x, we have x = F/k. Plugging in the values given, x = 430 N / 700 N/m = 0.614 m. Therefore, the spring will stretch 0.614 meters.
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10.
What is the equation for frequency?
A.
F = 1/T
B.
F = t/n
C.
F = 2 T
D.
F= v/t
Correct Answer
A. F = 1/T
Explanation The equation for frequency is given by f = 1/T, where f represents frequency and T represents the period. This equation states that frequency is equal to the reciprocal of the period. In other words, frequency is the number of cycles or oscillations that occur in a unit of time.
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