A block weighing 4.0 kg extends a spring by 0.16 m from its upstretched position. The block is removed and a 0.50 kg body is hung from the same spring. If the spring is now stretched and then released, what is its period of vibration? (T = 0.28 )


Given:

Mass of the block = m₁  = 4.0 kg
displacement = x = 0.16 m

Mass of the body = m₂  = 0.50 kg

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

To Find:


The time period of the vibration due to mass m₂  = T = ?


Solution:


The general formula for the Time Period of a mass attached to a spring is 

T = 2Ï€ `\sqrt frac{m}{k}`

Here we have to find first the value of the spring constant due to mass of the block m₁. So by using Hook's Law

F = k x

or

k = `\frac {F}{x}` = `\frac {mg}{x}`

by putting values

k = `\frac {4 kg ｘ 9.8 m s⁻²}{0.16 m}`

k =  245 N m⁻¹ 


Now due to the vibration of the body of mass m₂, the Time period is 

T = 2Ï€ `\sqrt frac{m₂}{k}`

by putting values 

T = 2Ï€ `\sqrt frac{0.50 kg}{245 N m⁻¹}`

T = 2 ｘ 3.1416 ｘ `sqrt 0.00204 s²`

T = 6.283 ï½˜ 0.045 s

T = 0.283 s ------------------Ans.




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