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  • How to solve this problem- - Oscillations and Waves - JEE Main-3

Two springs, of force constants k_{1} and k_{2} are connected to a mass m as shown. The frequency of oscillation of the mass is f.      If both k_{1} and k_{2} are made four times their original values, the frequency of oscillation becomes

  • Option 1)

    2f

  • Option 2)

    f/2

  • Option 3)

    f/4

  • Option 4)

    4f

 

Answers (1)

best_answer

As we learnt in

Parallel combination of spring -

- wherein

K_{eq}=K_{1}+K_{2}

K_{1}and\ K_{2} are spring constants of spring 1 & 2 respectively.

 

 

 

In the given figure two springs are connected in parallel. Therefore the effective spring constan is given by

k_{eff}= k_{1}+k_{2}                

Frequency of oscillation,

f= \frac{1}{2\pi }\sqrt{\frac{k_{eff}}{m}}= \frac{1}{2\pi }\sqrt{\frac{k_{1}+k_{2}}{m}}\cdots \cdots \cdots \cdots (i)

As    k_{1}   and   k_{1}   are increased four times N ew frequency,

{f}'= \frac{1}{2\pi }\sqrt{\frac{4\left ( k_{1}+k_{2} \right )}{m}}= 2f\: \: \: (Using (i))

Correct option is 1.


Option 1)

2f

This is the correct option.

Option 2)

f/2

This is an incorrect option.

Option 3)

f/4

This is an incorrect option.

Option 4)

4f

This is an incorrect option.

Posted by

Plabita

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