Question

A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period of 1.50 s. When the 0.750-kg object is removed and replaced by a 1.50-kg object, what will be the period of oscillation

Answer 1

New time period,
`T_2=2.12\ s`

Step-by-step explanation:

Given that,

Mass of the object 1,
`m_1=0.75\ kg`

Time period,
`T_1=1.5\ s`

If object 1 is replaced by object 2,
`m_2=1.5\ kg`

Let
`T_2` is the new period of oscillation.

The time period of oscillation of mass 1 is given by :

`T_1=2\pi \sqrt{(m_1)/(k)}`

`1.5=2\pi \sqrt{(0.75)/(k)}`............(1)

The time period of oscillation of mass 2 is given by :

`T_2=2\pi \sqrt{(m_2)/(k)}`

`T_2=2\pi \sqrt{(1.5)/(k)}`............(2)

From equation (1) and (2) we get :

`((T_1)/(T_2))^2=(m_1)/(m_2)`

`((1.5)/(T_2))^2=(0.75)/(1.5)`

`(1.5)/(T_2)=0.707`

`T_2=2.12\ s`

So, the new period of oscillation is 2.12 seconds. Hence, this is the required solution.