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

Answer:

New time period, $T_2=2.12\ s$

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 :

$T_1=2\pi \sqrt{\dfrac{m_1}{k}}$

$1.5=2\pi \sqrt{\dfrac{0.75}{k}}$............(1)

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

$T_2=2\pi \sqrt{\dfrac{m_2}{k}}$

$T_2=2\pi \sqrt{\dfrac{1.5}{k}}$............(2)

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

$(\dfrac{T_1}{T_2})^2=\dfrac{m_1}{m_2}$

$(\dfrac{1.5}{T_2})^2=\dfrac{0.75}{1.5}$

$\dfrac{1.5}{T_2}=0.707$

$T_2=2.12\ s$

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