A simple pendulum is 50 cm long. What will be its frequency of vibration at a place where g = 9.8 m s⁻². (Ans: 0.70 Hz)



Data Given:


Length of the pendulum = L = 50 cm = 0.5 m

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

To Find:


Frequency of the simple pendulum  = f = ?


Solution:

The general formula for Time Period of a simple pendulum is

T = 2π `\sqrt frac{L}{g}`

but we know the frequency and time period relationship is 

f = `\frac {1}{T}`

thus general formula for frequency of a simple pendulum will be

f = `\frac {1}{2π}` `\sqrt frac{g}{l}`

by putting values 

f = `\frac {1}{2 ｘ 3.1416}` `\sqrt frac{9.8 m s⁻²}{0.5 m}`

f = `\frac {1}{6.292}` ｘ `sqrt 19.6 s⁻²`

f = 0.159ｘ 4.427 s⁻¹

f = 0.704 Hz   {s⁻¹ = Hz}

or

 f = 0.7 Hz  ------------------Ans.



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