A simple pendulum completes one oscillation in 4 seconds. Calculate its length when g= 9.8 m/s2, as the time period of a simple pendulum is given by T = 2 π √(l/g)
Correct answer: A. 3.973 m
- A. 3.973 m
- B. 5.123 m
- C. 7.111 m
- D. 9.231 m
- E. 12.141 m
Explanation
To find the length of the pendulum, use the formula for the time period of a simple pendulum: T = 2π√(l/g). Given that T = 4 s and g = 9.8 m/s², rearrange the formula to solve for l:l = (T²g)/(4π²).Substitute the given values:l = (4² × 9.8)/(4 × π²) ≈ 3.973 m.Therefore, the correct length is approximately 3.973 meters. Other options result from errors in rearranging the formula or incorrect substitution of values.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
Practise Waves
1,281 free Waves MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Simple Harmonic Motion questions
When there is an increase in the length of the simple pendulum then its time period will:
A body of mass 4 kg attached to a spring is displaced through 0.04 m from its equilibrium position and then released. If the spring constant is 400 N/m, find the time period of the vibration?
A student is performing a lab experiment on simple harmonic motion. He has two different springs (with force constants k1 and k2) and two different blocks (of masses m1 and m2). If k1 =2k2, and m1 = 2m2, which of the following combinations would give the student the spring-block simple harmonic oscillator with the shortest period?
Two springs fixed at one end are stretched by 5 cm and 10 cm, respectively, when masses 0.5 kg and 1 kg are suspended at their lower ends. When displaced slightly from their mean positions and released, they will oscillate with time periods in the ratio
The time of open vibration of a simple pendulum may be decreased by:
At which of the following places, motion of simple pendulum becomes slowest: