Question

A system is executing S.H.M. with a periodic time of \(4 / 5 \mathrm{~s}\) under the influence of force \(\mathrm{F}_{1}\) When a force \(\mathrm{F}_{2}\) is applied, the periodic time is \((2 / 5) \mathrm{s}\). Now if \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) are applied simultaneously along the same direction, the periodic time will be......... (A) \(\\{4 /(5 \sqrt{5})\\}\) (B) \(\\{4 /(4 \sqrt{5})\\}\) (C) \(\\{8 /(4 \sqrt{5})\\}\) (D) \(\\{8 /(5 \sqrt{5})\\}\)

Short answer

\(T_{sim} = \frac{4}{5\sqrt{5}} s\)

Step-by-step solution

Determine the time periods for F1 and F2 individually

We are given the following periodic times for the system under the influence of forces F1 and F2: - Periodic time under force F1 is \(T_1 = \frac{4}{5} s\) - Periodic time under force F2 is \(T_2 = \frac{2}{5} s\)

Apply the formula for the periodic time

The periodic time \(T\) of an SHM is given by the formula: \[T = 2\pi \sqrt{\frac{m}{k}}\] Where - \(T\) is the periodic time - \(m\) is the mass of the system - \(k\) is the force constant We will now apply this formula for the periods \(T_1\) and \(T_2\).

Write the equations for periods under forces F1 and F2

We have the following equations for the periods under the two different forces: \[T_1 = 2\pi \sqrt{\frac{m}{k_1}}\] \[T_2 = 2\pi \sqrt{\frac{m}{k_2}}\] Where \(k_1\) and \(k_2\) are the force constants under the influence of forces F1 and F2, respectively.

Create an equation relating the forces and the periodic times

First, square both equations to get rid of the square root sign: \[T_1^2 = 4\pi^2 \frac{m}{k_1}\] \[T_2^2 = 4\pi^2 \frac{m}{k_2}\] Next, divide the first equation by the second equation: \[\frac{T_1^2}{T_2^2} = \frac{4\pi^2 \frac{m}{k_1}}{4\pi^2 \frac{m}{k_2}}\] Simplify the equation to relate the force constants: \[\frac{k_2}{k_1} = \frac{T_1^2}{T_2^2}\]

Calculate the force constant ratio

Substitute the given time periods for \(T_1\) and \(T_2\) and compute \(\frac{k_2}{k_1}\): \[\frac{k_2}{k_1} = \frac{(\frac{4}{5})^2}{(\frac{2}{5})^2} = \frac{16}{4} = 4\]

Determine the new force constant

Now, both force constants act on the system simultaneously along the same direction. The new force constant, \(k_{sim}\), is the sum of the individual force constants: \[k_{sim} = k_1 + k_2 = k_1 + 4k_1 = 5k_1\]

Determine the new periodic time

Now we can find the new periodic time, \(T_{sim}\), using the following equation: \[T_{sim} = 2\pi \sqrt{\frac{m}{k_{sim}}}\] We know that \(T_1 = 2\pi \sqrt{\frac{m}{k_1}}\), thus, we can solve for \(\sqrt{\frac{m}{k_{1}}}\) to find: \[\sqrt{\frac{m}{k_{1}}} = \frac{T_1}{2\pi}\] Now, substitute \(\sqrt{\frac{m}{k_{1}}}\) and \(k_{sim} = 5k_1\) into the \(T_{sim}\) equation and simplify: \[T_{sim} = 2\pi \sqrt{\frac{m}{5k_{1}}} = 2\pi (\frac{1}{\sqrt{5}})\frac{T_1}{2\pi}\] After simplification, we get the new periodic time \(T_{sim}\) as: \[T_{sim} = \frac{4}{5\sqrt{5}} s\] So, the correct option is: (A) \(\frac{4}{5\sqrt{5}}\)

Key concepts

Periodic Time

In Simple Harmonic Motion (SHM), "Periodic Time" is an essential concept. It refers to the time taken by an object to complete one full cycle of the motion. In this exercise, the periodic time for two forces, \(F_1\) and \(F_2\), are given as \(\frac{4}{5}\) seconds and \(\frac{2}{5}\) seconds respectively. Periodic time is determined by the mass of the system and the force constant, which measures the system's stiffness. The equation used to find the periodic time \(T\) is:\[T = 2\pi \sqrt{\frac{m}{k}}\]Here, \(m\) is the mass and \(k\) is the force constant. A smaller periodic time means the system oscillates faster compared to a larger periodic time. Understanding periodic time helps predict how speedy or slow a system returns to equilibrium.

Force Constant

The "Force Constant" is a crucial part of understanding SHM. It's denoted by \(k\), and it tells us how stiff or strong the restoring force is for the system. You can think of it as the springiness. In the exercise, each force \(F_1\) and \(F_2\) influences the force constant differently, represented as \(k_1\) and \(k_2\). The formula connects the force constant with periodic time:\[T = 2\pi \sqrt{\frac{m}{k}}\]Where a higher force constant means a stiffer spring which results in faster oscillations (shorter \(T\)). Calculating \(\frac{k_2}{k_1}\) here shows the relationship between the two different conditions of force application. When additional force is applied, the effective force constant becomes the sum of the individual constants, resulting in an overall stiffer system.

Mass

Mass plays a fundamental role in the periodic motion of systems like springs and pendulums in SHM. The mass \(m\) of the system impacts how long it takes for the system to complete a cycle, as shown in the equation:\[T = 2\pi \sqrt{\frac{m}{k}}\]In our exercise, though the mass is constant for the forces \(F_1\) and \(F_2\), it affects how the system responds to these forces. A more massive object tends to oscillate more slowly, resulting in a longer periodic time due to its higher inertia. Thus, mass, along with force constant, helps determine the dynamics of SHM.

Simultaneous Forces

When forces \(F_1\) and \(F_2\) are applied simultaneously, they influence the system's behaviour significantly through their combined impact on the force constant. In the problem at hand, both forces are applied in the same direction, making their effects cumulative. Therefore, the combined force constant \(k_{sim}\) becomes\[k_{sim} = k_1 + k_2\]This results in a different periodic time, as calculated by:\[T_{sim} = 2\pi \sqrt{\frac{m}{k_{sim}}}\]The simultaneous application leads to a stiffer system due to an increased force constant from both forces, altering the motion's speed and frequency. This concept highlights the importance of understanding how multiple forces interact to shape motion in SHM.