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

The frequency of oscillation for a mass connected to springs is given by the formula:

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

Here, $$k_{eq}$$ represents the equivalent spring constant of the system. Whether the springs are connected in series or in parallel, the equivalent spring constant is a linear combination of $$k_1$ and $k_2$$.

Let the initial equivalent spring constant be $$k_{eq}$$. When both $$k_1$$ and $$k_2$$ are made four times their original values, the new equivalent spring constant becomes:

$$k_{eq}' = 4k_{eq}$$

Now, substituting this into the frequency formula to find the new frequency $$f'$$:

$$f' = \frac{1}{2\pi} \sqrt{\frac{4k_{eq}}{m}}$$

Taking the square root of 4 outside the radical:

$$f' = 2 \times \left(\frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}}\right)$$

$$f' = 2f$$