Question

As shown in figure, a body having mass m is attached
with two springs having spring constants k, and k,
The frequency of oscillation is f. Now, if the springs
constants of both the springs are increased 4 times, then
the frequency of oscillation will be equal to
ky
k 2
sooooo sooooo
f
(A) 2f
(B)
f
2
(D) 4f
(C)
4​

Answer 1

Correct Question:

As shown in figure, a body of mass m is attached with 2 springs having spring constants k1 and k2. The frequency of oscillation is f. Now, when the springs constants of both the springs is increased 4 times, then the frequency of oscillation will be equal to?

Calculation:

This type of attachment of springs are same as that of parallel combination, such that :

Initial spring constant = k1 + k2

So, initial frequency :

$\therefore f1 = \dfrac{1}{2\pi} \sqrt{ \dfrac{k_{net} }{m} }$

$= > f1 = \dfrac{1}{2\pi} \sqrt{ \dfrac{k1 + k2 }{m} }$

Now , spring constants of both the spring is increased by 4 times , hence the final net spring constant:

k_(net) = 4(k1) + 4(k2) = 4 (k1 + k2)

So, final frequency:

$\therefore f2 = \dfrac{1}{2\pi} \sqrt{ \dfrac{k_{net} }{m} }$

$= > f2 = \dfrac{1}{2\pi} \sqrt{ \dfrac{4(k1 + k2) }{m} }$

$= > f2 = 2 \times \bigg (\dfrac{1}{2\pi} \sqrt{ \dfrac{k1 + k2 }{m} } \bigg)$

$= > f2 = 2 \times \bigg (f\bigg)$

$= > f2 = 2f$

So, final frequency of combination of springs is 2f.

Hope It Helps.