A simple pendulum of length 1.5m has a bob of mass 2.0kg. State the formula for the period of small oscillations and evaluate it

Question

2 years ago

5

in this case. (2.43s)

1 answer:

Shalnov [3] · Answer 1 · 2024-01-26T04:24:44+03:00

A simple pendulum of length 1.5m has a bob of mass of 2.0kg, then the period for the small oscillations would be 2.456 s.

<h3>What is the frequency?</h3>

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,

f = 1/T

As given in the problem A simple pendulum of length 1.5m has a bob of mass of 2.0kg.

The formula for the time period for the pendulum,

T = 2π√L/g

=2π√1.5/9.81

=2.456 s

Thus, the period for the small oscillations would be 2.456 s.

Learn more about frequency from here,

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