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BITSAT 2019 Paper

Section: Physics

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Question : 22 of 150

Marks: +1, -0

Two springs, of force constants k

{}_1

{}_2

are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k

{}_1

{}_2

are made four times their original values, the frequency of oscillation becomes

2f
f/2
f/4
4f

Solution:

(a) The two springs are in parallel.

∴ Effective spring constant,

k =

k_1+k_2

Now, frequency of oscillation is given by

f =

\frac{1}{2p}\sqrt{\frac{k}{m}}

or, f =

\frac{1}{2p}\sqrt{\frac{k_1+k_2}{m}}

........(i)

When both

k_1

and

k_2

are made four timestheir original values, the new frequency is given by

f'

=

\frac{1}{2p}\sqrt{\frac{4k_1+4k_2}{m}}

f'

=

\frac{1}{2\pi}\sqrt{\frac{4(k_1+ k_2)}{m}}

=

2\left(\frac{1}{2\pi}\sqrt{\frac{ (k_1+ k_2)}{m}}\right)

= 2f

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