Simple pendulum and dissipation of energy — practice questions
10 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.
For small oscillations, the time period of a simple pendulum is approximately
- A. T=2πsqrt(g/L)
- B. T=L/g
- C. T=2πsqrt(L/g)
- D. T=2πL/g
The period of a simple pendulum is approximately independent of
- A. pendulum length
- B. local g
- C. large amplitude outside the small-angle limit
- D. bob mass
If pendulum length is quadrupled, its small-angle period becomes
- A. twice
- B. four times
- C. half
- D. unchanged
To reduce timing reaction error in a pendulum experiment, one should time
- A. only a fraction of one oscillation
- B. many oscillations and divide by their number
- C. one oscillation repeatedly without averaging
- D. the bob mass
Damping causes pendulum amplitude to
- A. increase indefinitely
- B. remain exactly constant in air
- C. decrease with time
- D. reverse sign permanently after each cycle
In a damped pendulum, lost mechanical energy is mainly converted into
- A. extra gravitational potential that accumulates forever
- B. mass
- C. electric charge
- D. thermal energy and sound in the surroundings
A graph of amplitude versus time for a lightly damped pendulum is expected to show
- A. a gradually decaying envelope
- B. constant amplitude forever
- C. linear growth without bound
- D. instantaneous zero after one cycle
If measured period is used to find g, the relation is
- A. g=T²/(4π²L)
- B. g=4π²L/T²
- C. g=2πL/T
- D. g=LT²
A student measures time for 20 oscillations as 40.0 s. The estimated period is
- A. 20.0 s
- B. 0.50 s
- C. 2.00 s
- D. 800 s
Why should the pendulum amplitude be kept small in the standard g experiment?
- A. To make bob mass zero
- B. To eliminate gravity
- C. To force period to depend strongly on amplitude
- D. So the small-angle approximation underlying T=2πsqrt(L/g) remains valid
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