Solved Numerical Problem No. 7 ( Unit-7: Oscillation); HSSC-1 (Class-11); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

What should be the length of a simple pendulum whose time period is one second? What is its frequency? ((1) = 0.25 m (ii) f = 1 cps)

Given:

The time period of the simple pendulum = T = 1 s

Value of acceleration due to gravity = g = 9.8 m s⁻²

To Find:

Length of second pendulum = L = ?

Frequency = f = ?

Solution:

The formula for the Time Period of a simple pendulum is 

T = 2π `\sqrt frac{L}{g}`

To separate the L from this formula we will take square on both sides and by simplifying we get

L = `\frac {T^2 g}{(2π)^2}`

by putting values

L = `\frac {(1s)^2 ｘ 9.8 m s⁻²}{(2 ｘ 3.1416)^2}`

 

L = `\frac {1 s² ｘ 9.8 m s⁻²}{(6.283185)^2}`

L = `\frac {9.8 m}{39.478}`

L = 0.248 m ------------Ans.(1)

As frequency f and time period T are related inversely ie.

f = `\frac {1}{T}` 

f = `\frac {1}{1 s}` 

f = 1 Hz 

or 

f = 1 cps------------Ans.(2)

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