Question

A mass m is attached to two springs of same force constants as shown in figure. What is T₁/T₂?
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

Given:

Mass m is attached to two springs of same force constants in two different cases as shown in the figure.

To find:

Ratio of time period of oscillation of blocks in the two cases.

Concept:

In both the setup , we need to find out the net Spring Constant . Then , we can calculate the time period of oscillation of the block.

Calculation:

In the 1st case :

The springs are attached in parallel combinations , such that the equivalent spring constant will be algebraic addition of individual spring constants

$\therefore \: k \: eq. = k1 + k2$

$= > k \: eq. = k + k$

$= > k \: eq. = 2k$

So time period will be :

$\therefore \: T_{1} = 2\pi \sqrt{ \dfrac{m}{k \: eq.} }$

$= > \: T_{1} = 2\pi \sqrt{ \dfrac{m}{2k} }$

In the 2nd case:

The springs are again connected to one another in parallel combination such that the net spring constant will again be the algebraic sum of the individual string constants.

$\therefore \: k \: eq. = k1 + k2$

$= > k \: eq. = k + k$

$= > k \: eq. = 2k$

So , time period will be :

$\therefore \: T_{2} = 2\pi \sqrt{ \dfrac{m}{k \: eq.} }$

$= > \: T_{2} = 2\pi \sqrt{ \dfrac{m}{2k} }$

So , require ratio will be :

$\therefore \: T_{1} :T_{2} = 2\pi \sqrt{ \dfrac{m}{2k} } : 2\pi \sqrt{ \dfrac{m}{2k} }$

$= > \: T_{1} :T_{2} = 1: 1$

So the final answer is :

T1/T2 = 1