Question

A block attached to a spring with unknown spring constant oscillates with a period of 7.0 s .What is the period if the mass is doubled

Answer 1

Answer:

The period is 9.9 seconds

Explanation:

The period (T) of a mass (m) attached to a spring with spring constant k is:

$T=2\pi\sqrt{\frac{m}{k}}$ (1)

So, if the mas is M we have that:

$7.0s=T_{1}=2\pi\sqrt{\frac{M}{k}}$ (2)

Now if we double the mass:

$T_{2}=2\pi\sqrt{\frac{2M}{k}}$

$T_{2}=\sqrt{2}(2\pi\sqrt{\frac{M}{k}})$ (3)

Because spring constant doesn’t change, we note that the term $2\pi\sqrt{\frac{M}{k}}$ on (3) is equal to the right side of (2), so we have:

$T_{2}=\sqrt{2}T_{1}=\sqrt{2}(7.0s)$

$T_{2}=9.9s$