Single Choice

The length of a simple pendulum is made one-fourth. Its time period becomes:

Afour times
Bone-fourth
Cdouble
Dhalf
Correct Answer

Solution

$$T=2\pi\sqrt{\dfrac{L}{g}}$$
If length is reduced to a fourth $$T'=2\pi\sqrt{\dfrac{L}{4g}}=\dfrac{2\pi}{2}\sqrt{\dfrac{L}{g}}=T/2$$

SIMILAR QUESTIONS

Measurement and Errors

Two simple pendulums A and B have equal lengths, but their bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? Give reason for your answer.

Measurement and Errors

Define the terms: (i) oscillation, (ii) amplitude (iii) frequency (iv) time period as related to a simple pendulum

Measurement and Errors

Name two factors on which the time period of a simple pendulum depends. Write the relation for the time period in terms of the above named factors.

Measurement and Errors

Name two factors on which the time period of a simple pendulum depends. Write the relation for the time period in terms of the above named factors.

Measurement and Errors

How does the time period (T) of a simple pendulum depend on its length(I)? Draw a graph showing the variation of $$T^2$$ with 1. How will you use this graph to determine the value of g (acceleration due to gravity)?

Measurement and Errors

A seconds' pendulum is taken to a place where acceleration due to gravity falls to one-forth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

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