Question

A simple pendulum is of length 50cm. Find its time period and frequency of oscillation
(8 = 9.8m/s(​

Answer 1

Answer:

time period of oscilation of pendulum=

$= 2\pi \sqrt{ \frac{r}{g} }$

=2π (√50cm/√9.8)

=2×3.14×(√0.5m/√9.8)

= 1.42 sec(approx)

now,

frequency = (time period)^–1

= 1/(1.42) = 0.704 s^–1

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Answer 2

Time period : 1.41 seconds

Frequency of oscillation, f = 0.709 Hz

Explanation:

Given that,

Length of the pendulum, l = 50 cm = 0.5 m

The time period of the simple pendulum is given by :

$T=2\pi\sqrt{\dfrac{l}{g}}$

$T=2\pi\sqrt{\dfrac{0.5}{9.8}}$

T = 1.41 seconds

If f is the frequency of oscillation. It is given by :

$f=\dfrac{1}{T}$

$f=\dfrac{1}{1.41}$

f = 0.709 Hz

So, the time period and the frequency of oscillation is 1.41 seconds and 0.709 Hz respectively.

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Topic : Simple pendulum

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