Question

In the fig. S1 and S2 are identical springs. The oscillation frequency of the mass m is f. If one spring is removed, the frequency will become
(a) f
(b) f × 2
(c) f × √2
(d) f/√2

Answer 1

answer : option (d) f/√2

explanation :

we know , time period of spring is given by, $T=2\pi\sqrt{\frac{m}{k_{eq}}}$

but we know, time period = 1/frequency

so, frequency, f = $\frac{1}{2\pi}\sqrt{\frac{k_{eq}}{m}}$

both the spring are connected in parallel combination.

so, $k_{eq}=k_1+k_2$

here, $k_1=k_2=k$

so, $k_{eq}=2k$

now, $f=\frac{1}{2\pi}\sqrt{\frac{2k}{m}}$.......(1)

if one spring is removed , $k_{eq}=k$

then, new frequency, $f'=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$......(2)

from equations (1) and (2),

$f'=\frac{f}{\sqrt{2}}$

hence, option (d) is correct choice.