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Chapter 10: Problem 1399

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

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

Step by step solution

01

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 \(T_2 = \frac{2}{5} s\)
02

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 \(T_2\).
03

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.
04

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: \[\frac{k_2}{k_1} = \frac{T_1^2}{T_2^2}\]
05

Calculate the force constant ratio

Substitute the given time periods for \(T_1\) and \(T_2\) and compute \(\frac{k_2}{k_1}\): \[\frac{k_2}{k_1} = \frac{(\frac{4}{5})^2}{(\frac{2}{5})^2} = \frac{16}{4} = 4\]
06

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: \[k_{sim} = k_1 + k_2 = k_1 + 4k_1 = 5k_1\]
07

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) \(\frac{4}{5\sqrt{5}}\)

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Most popular questions from this chapter

An open organ pipe has fundamental frequency \(100 \mathrm{~Hz}\). What frequency will be produced if its one end is closed? (A) \(100,200,300, \ldots\) (B) \(50,150,250 \ldots .\) (C) \(50,100,200,300 \ldots \ldots\) (D) \(50,100,150,200 \ldots \ldots\)

The maximum velocity and maximum acceleration of a particle executing S.H.M. are \(1 \mathrm{~m} / \mathrm{s}\) and \(3.14 \mathrm{~m} / \mathrm{s}^{2}\) respectively. The frequency of oscillation for this particle is...... (A) \(0.5 \mathrm{~s}^{-1}\) (B) \(3.14 \mathrm{~s}^{-1}\) (C) \(0.25 \mathrm{~s}^{-1}\) (D) \(2 \mathrm{~s}^{-1}\)

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