A mass M is suspended from a spring - AQA - A-Level Physics - Question 27 - 2022 - Paper 1

Initially, the elastic potential energy when mass M is at the equilibrium position is given as E. When an extra mass of 2M is added, the total mass becomes 3M.

Using Hooke's law, we know the elastic potential energy (E) stored in a spring is given by: $E = \frac{1}{2} k x^2$ where k is the spring constant and x is the extension.

When the mass is increased to 3M, the new elastic potential energy will be: $E_{new} = \frac{1}{2} k (x_1 + \Delta x)^2$ Where \Delta x is the additional extension caused by the extra mass. Considering that the potential energy depends on the square of the extension, we can conclude that:

If we look at it simply, the new energy stored, given the conditions remains proportional to the total mass, we can deduce:

• The initial energy ($E$) was proportional to M.
• The new energy will thus be proportional to 3M, and mathematically we can say it can be expressed as:

$E_{new} = 3E$

Therefore, the total elastic energy stored in the system when it is at rest at the new equilibrium position is 3E.