Question

When a mass \(M\) is suspended from the free end of a spring, its periodic time is found to be \(\mathrm{T}\). Now, if the spring is divided into two equal parts and the same mass \(\mathrm{M}\) is suspended and oscillated, the periodic time of oscillation is found to be \(\mathrm{T}\) '. Then \(\ldots \ldots \ldots\) (A) \(\mathrm{T}<\mathrm{T}^{\prime}\) (B) \(\mathrm{T}=\mathrm{T}^{\prime}\) (C) \(\mathrm{T}>\mathrm{T}^{\prime}\) (D) Nothing can be said.

Short answer

(A) \(T < T'\)

Step-by-step solution

Write down the formula for the periodic time

The formula for the periodic time of a mass-spring system is given by \(T = 2\pi\sqrt{\frac{M}{k}}\) where \(T\) is the periodic time, \(M\) is the mass, and \(k\) is the spring constant of the spring.

Determine the relation between periodic time and spring constant

We are given that when the spring is divided into two equal parts, the mass M will oscillate with a new periodic time T'. Let's denote the spring constant of the original undivided spring as \(k_1\) and the spring constant of the divided springs as \(k_2\). We will use the above formula for the periodic time with these two spring constants to obtain: \(T = 2\pi\sqrt{\frac{M}{k_1}}\) and \(T' = 2\pi\sqrt{\frac{M}{k_2}}\)

Relate the spring constants of the divided and undivided spring

We know that when we divide the spring into two equal parts, the spring constant is also changed. For a spring, the spring constant is inversely proportional to its length (\(k \propto \frac{1}{L}\)). Since the spring is divided into two equal parts, the length of each part is half the length of the original spring. Thus, the spring constant of the divided springs, \(k_2\), is twice the spring constant of the original spring: \(k_2 = 2k_1\)

Compare the periodic times T and T'

Now we can use this relationship between the spring constants to compare the periodic times T and T'. By plugging the expression for \(k_2\) in terms of \(k_1\) from Step 3 into the T' equation from Step 2, we can obtain: \(T' = 2\pi\sqrt{\frac{M}{2k_1}}\) Now, dividing the T' equation by the original T equation: \(\frac{T'}{T} = \frac{2\pi\sqrt{\frac{M}{2k_1}}}{2\pi\sqrt{\frac{M}{k_1}}} = \sqrt{\frac{1}{2}}\) Since \(0 < \sqrt{\frac{1}{2}} < 1\), we have: \(T' < T\) So, the correct answer is: (A) \(T < T'\)