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) \(T < T'\)

Key concepts

Spring-Mass System

The spring-mass system is a classic physics model used to study oscillations. It involves a spring attached to a mass that can move back and forth. This setup is not just for clocks or toys; it actually represents any system where a constant force (like a spring) affects a movable object (the mass). You will find it in places from car suspensions to seismic vibration absorbers.
In the system, the spring provides a restorative force to the mass, encouraging it to oscillate. These oscillations happen because the spring tries to return to its original shape, pulling back the mass attached to it. The back-and-forth motion continues unless an external force stops it.
Key factors in this system are the spring constant, which measures the stiffness of the spring, and the mass attached to it. A higher spring constant results in a stiffer spring, while a larger mass results in slower oscillations. Understanding this concept helps in designing systems where controlled oscillation is crucial.

Periodic Time

Periodic time, often called the period, is the time it takes to complete one full cycle of oscillation in a spring-mass system. Imagine a mass attached to a spring; periodic time is the duration for it to start at one point, return to that same point, and complete the loop.
The formula to calculate periodic time, given by \(T = 2\pi\sqrt{\frac{M}{k}}\), shows that periodic time depends on the mass \(M\) and the spring constant \(k\). This formula tells us:

• The larger the mass, the longer the periodic time, meaning the oscillations are slower.
• The stiffer the spring (larger spring constant), the shorter the periodic time, which makes the oscillations quicker.

Periodic time is crucial in many real-world applications such as designing pendulum clocks and even systems in engineering where timing is important. By adjusting mass or spring constant, one can fine-tune the period according to specific needs.

Spring Constant

The spring constant, denoted as \(k\), is a measure of a spring's stiffness. It's a key characteristic that tells you how much force you need to stretch or compress the spring by a unit length. In essence, it defines how "firm" or "loose" a spring is.
Hooke's Law, which states \(F = kx\), ties in the spring constant with force \(F\) and displacement \(x\). More force is required to stretch or compress a stiffer spring a certain distance. Therefore, a spring with a large constant is less easily deformed, meaning it takes more force to change its length compared to a spring with a smaller constant.
When a spring is divided into two equal parts, its length decreases, thus increasing the spring constant. This is because the spring constant is inversely proportional to length. Specifically, if the original spring's constant was \(k_1\), cutting it in half results in a constant \(k_2 = 2k_1\). Changes in the spring constant influence the system's oscillation behavior and are crucial in designing systems requiring precise motion control, like in shock absorbers or mechanical watches.