Question

A mass m is attached to an ideal massless spring. When this system is set in motion, it has a period t. What is the period if the mass is doubled to 2m?.

Answer 1

If a mass m is attached to an ideal massless spring and has a period of t, then the period of the system when the mass is 2m is $\sqrt{2}t$.

Calculation:

Step-1:

It is given that a mass m is attached to an ideal massless spring and the period of the system is t. It is required to find the period when the mass is doubled.

The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T.

It is known that the period is calculated as,

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

Here m is the mass of the object, and k is the spring constant.

Step-2:

Thus the period of the system with the first mass is,

$t=2 \pi \sqrt{\frac{m}{k}}$

The period of the system with the second mass is,

$\begin{aligned}\\t^'&=2 \pi \sqrt{\frac{m}{k}}\\&=\sqrt{2}\times2 \pi \sqrt{\frac{2m}{k}}\\&=\sqrt{2}\times t\end{aligned}$

Then the period of the system with the second mass is $\sqrt{2}$ times more than the period of the system with the first mass.

Learn more about period of a spring-mass system here,

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