Step 1: Recall the formula for natural frequency.
The natural frequency \( f \) of a spring-mass system is given by:
\[
f = \frac{1}{2\pi} \sqrt{\frac{k}{m}},
\]
where \( k \) is the spring constant, and \( m \) is the mass. The frequency in Hz is related to the angular frequency \( \omega \):
\[
\omega = 2\pi f = \sqrt{\frac{k}{m}}.
\]
Step 2: Set up equations for the two cases.
Case 1: Original mass \( m \), natural frequency \( f_1 = 2 \, \text{Hz} \):
\[
\omega_1 = 2\pi f_1 = 2\pi \times 2 = 4\pi,
\]
\[
\omega_1 = \sqrt{\frac{k}{m}},
\]
\[
4\pi = \sqrt{\frac{k}{m}},
\]
\[
(4\pi)^2 = \frac{k}{m},
\]
\[
16\pi^2 = \frac{k}{m} \quad (1).
\]
Case 2: Mass \( m + 1 \), natural frequency \( f_2 = 1 \, \text{Hz} \):
\[
\omega_2 = 2\pi f_2 = 2\pi \times 1 = 2\pi,
\]
\[
\omega_2 = \sqrt{\frac{k}{m + 1}},
\]
\[
2\pi = \sqrt{\frac{k}{m + 1}},
\]
\[
(2\pi)^2 = \frac{k}{m + 1},
\]
\[
4\pi^2 = \frac{k}{m + 1} \quad (2).
\]
Step 3: Solve for \( m \).
Divide equation (1) by equation (2):
\[
\frac{16\pi^2}{4\pi^2} = \frac{\frac{k}{m}}{\frac{k}{m + 1}},
\]
\[
4 = \frac{m + 1}{m},
\]
\[
4m = m + 1,
\]
\[
3m = 1,
\]
\[
m = \frac{1}{3} \, \text{kg}.
\]
Step 4: Verify the solution.
With \( m = \frac{1}{3} \, \text{kg} \):
From (1):
\[
16\pi^2 = \frac{k}{\frac{1}{3}},
\]
\[
k = 16\pi^2 \times \frac{1}{3} = \frac{16\pi^2}{3}.
\]
From (2):
\[
m + 1 = \frac{1}{3} + 1 = \frac{4}{3},
\]
\[
4\pi^2 = \frac{k}{\frac{4}{3}},
\]
\[
k = 4\pi^2 \times \frac{4}{3} = \frac{16\pi^2}{3},
\]
which matches. The value of \( m \) is consistent.
Step 5: Select the correct answer.
The original mass \( m \) is \( \frac{1}{3} \) kg, matching option (3).