The time period of oscillation of a mass-spring system depends on the mass of the object attached to the spring and the force constant of the spring. In this case, the spring has been cut into four equal parts, and one part is attached to a mass.

1. Force constant of the spring:
The force constant (k) of the spring remains constant regardless of whether the spring is cut into equal parts or not. The force constant represents the stiffness of the spring and is given by Hooke's Law as the ratio of the force applied to the displacement produced. It is denoted by the symbol k.

2. Calculation of new force constant:
When the spring is cut into four equal parts, each part will have a force constant of k/4. This is because the force constant is directly proportional to the length of the spring. So, if the length of the spring is divided into four equal parts, each part will have a force constant that is one-fourth of the original force constant.

3. Effect on time period of oscillation:
The time period of oscillation (T) of a mass-spring system is given by the formula T = 2π√(m/k), where m is the mass attached to the spring and k is the force constant of the spring.

When the mass is attached to one part of the spring, the effective force constant becomes k/4. Plugging this value into the time period formula, we get T = 2π√(m/(k/4)). Simplifying this equation, we get T = 2π√(4m/k), which can be further simplified to T = 2π√(m/k).

4. Conclusion:
From the above calculation, we can see that cutting the spring into four equal parts and attaching one part to a mass does not affect the time period of oscillation. The time period remains the same and is given by T = 2π√(m/k). This means that the mass-spring system will oscillate with the same frequency and time period irrespective of whether the spring is cut or not.