Question

$\text{In a spring mass vibrating system, the natural frequency of vibration is reduced to half the value when a second spring is added to the first spring in series. The stiffness}\left(k_2\right)\text{of the second spring in terms of that of first spring}\left(k_1\right)\text{is}$

• A. $k_2=\frac{k_1}{2}$
• B. $k_2=\frac{k_1}{3}$
• C. $k_2=2k_1$
• D. $k_2=3k_1$

Answer 1

The correct option is B $k_2=\frac{k_1}{3}$

$f_n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

If frequency is to be halved, the stiffness must be reduced to one fourth.

$\text{Let the stiffness of the second spring be x time k}(k_2=xk)$

The combined stiffness k will be,

$\frac{1}{k^{\prime }}=\frac{1}{k_1}+\frac{1}{k_2}$

$\frac{1}{k^{\prime }}=\frac{1}{k}+\frac{1}{xk}$

$\therefore \frac{1}{\frac{k}{4}}=\frac{k(x+1)}{x{k}^2}$

$\therefore 4=\frac{x+1}{x}$

$\therefore x=\frac{1}{3}$

$\therefore k_2=\frac{1}{3}{k}_1$

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