The period of simple harmonic motion (SHM) is defined as: \[ T = 2\pi \sqrt{\frac{m}{k}}, \] where \( T \) represents the time period, \( m \) denotes the mass, and \( k \) is the spring constant.
Step 1: Determine the initial time period.
The time period associated with mass \( m \) is given by: \[ T_1 = 2 = 2\pi \sqrt{\frac{m}{k}}. \] Squaring both sides yields: \[ T_1^2 = \frac{4\pi^2 m}{k}. \] Rearranging the equation to solve for \( m \): \[ m = \frac{k T_1^2}{4\pi^2}. \]
Step 2: Calculate the new time period with the added mass.
The new time period is observed to be \( T_2 = 5 \, \mathrm{s} \), and the total mass increases to \( m + 0.04 \): \[ T_2^2 = \frac{4\pi^2 (m + 0.04)}{k}. \] Rearranging this equation: \[ m + 0.04 = \frac{k T_2^2}{4\pi^2}. \]
Step 3: Subtract the initial equation from the new equation.
Using the two derived equations: \[ \frac{k T_2^2}{4\pi^2} - \frac{k T_1^2}{4\pi^2} = 0.04. \] Factoring out common terms: \[ \frac{k}{4\pi^2} (T_2^2 - T_1^2) = 0.04. \] Substituting the given values of \( T_1 = 2 \, \mathrm{s} \) and \( T_2 = 5 \, \mathrm{s} \): \[ \frac{k}{4\pi^2} (5^2 - 2^2) = 0.04. \] Simplifying the expression: \[ \frac{k}{4\pi^2} \cdot (25 - 4) = 0.04. \] \[ \frac{k}{4\pi^2} \cdot 21 = 0.04. \] Solving for the ratio \( \frac{k}{4\pi^2} \): \[ \frac{k}{4\pi^2} = \frac{0.04}{21}. \]
Step 4: Substitute the found ratio back into the equation for \( m \).
Utilizing the expression for \( m \): \[ m = \frac{k T_1^2}{4\pi^2}. \] Substituting \( T_1 = 2 \, \mathrm{s} \) and the calculated value of \( \frac{k}{4\pi^2} \): \[ m = \left(\frac{0.04}{21}\right) \cdot 4. \] Simplifying the calculation: \[ m = \frac{0.04 \cdot 4}{21} = 0.00761 \, \mathrm{kg}. \] Converting the mass to grams: \[ m = 7.64 \, \mathrm{g}. \] Therefore, the calculated mass is \( \mathbf{7.64 \, \mathrm{g}} \).