Question

A disc of mass m is connected to two springs k1 and k2 as shown find time period of oscillation

Answer 1

heya mate!!

The frequency (f) of an oscillation is measure in hertz (Hz) it is the number of oscillations per second. The time for one oscillation is called the period (T) it is measured in seconds. Acceleration – we can calculate the acceleration of the object at any point in it's oscillation using the equation below.

hope it helps u dear✌️✌️

Answer 2

Given that,

Mass of disc = m

Spring constant is k₁ and k₂

Time period :

Time period of spring pendulum is defined as

$T=2\pi\sqrt{\dfrac{m}{k}}$

Where, m = mass of disc

k = spring constant

(I). If the spring constant k₁ and k₂ are connected in series

Then the effective spring constant will be

$\dfrac{1}{k}=\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}}$

$k=\dfrac{k_{1}k_{2}}{k_{1}+k_{2}}$

We need to calculate the time period of oscillation

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value of k in to the formula

$T=2\pi\sqrt{\dfrac{m}{\dfrac{k_{1}k_{1}}{k_{1}+k_{2}}}}$

$T=2\pi\sqrt{\dfrac{m(k_{1}+k_{2})}{k_{1}k_{2}}}$

(II). If the spring constant k₁ and k₂ are connected in parallel

Then the effective spring constant will be

$k=k_{1}+k_{2}$

We need to calculate the time period of oscillation

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value of k in to the formula

$T=2\pi \sqrt{\dfrac{m}{k_{1}+k_{2}}}$

Hence, (I). The time period of oscillation is $2\pi\sqrt{\dfrac{m(k_{1}+k_{2})}{k_{1}k_{2}}}$

(II). The time period of oscillation is $2\pi \sqrt{\dfrac{m}{k_{1}+k_{2}}}$