Question

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

• A. 2 f
• B. f/2
• C. f/4
• D. 4 f

Answer 1

Answer: (A)

Step-(i) Calculation of frequency, when spring constant Changes
Since, As a given - spring constant of both springs are made four times their original value
So $k_{1}'=4 k_{1}$
$k_{2} =4 k_{2}$
Now $k'=k_{1}'+k_{2}'=4 k_{1}+4 k_{2}$
$k'=4\left(k_{1}+k_{2}\right)$
Now frequency is given as -
$f'=\frac{1}{2 n} \sqrt{\frac{k'}{m}}=\frac{1}{2 \pi} \sqrt{\frac{4\left(k_{1}+k_{2}\right)}{m}}$
$f'=2\left(\frac{1}{2 n} \sqrt{\frac{k_{1}+k_{2}}{m}}\right)$ ...(ii)
from eq (i) & (ii)
$f'=2 f$