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Chapter 9: Problem 17

On smooth inclined plane, a body of mass \(m\) is attached between two massless springs. The other ends of the springs are fixed to firm supports. If each spring has force constant \(k\), the period of oscillation of the body is (A) \(2 \pi \sqrt{\frac{m}{2 k}}\) (B) \(2 \pi \sqrt{\frac{2 m}{k}}\) (C) \(2 \pi \sqrt{\frac{m g \sin \theta}{2 k}}\) (D) \(2 \pi \sqrt{\frac{2 m g \sin \theta}{k}}\)

Short Answer

Expert verified
The short answer is: The period of oscillation of the body on the smooth inclined plane is (A) \(2 \pi \sqrt{\frac{m}{2k}}\).

Step by step solution

01

Draw a free body diagram of the system

First, we should draw a free body diagram of the system, identifying all relevant forces acting on the mass 'm'. Along the inclined plane, there are three forces: gravitational force (mg), combined spring force (2kx), and tension forces. Note that the tension forces are not relevant for determining the motion, as they don't have components along the inclined plane.
02

Write down the equation of motion

Using Newton's second law, we can write down the equation of motion for the mass 'm' along the inclined plane. The net force acting on the mass equals mass times acceleration. Thus, the equation of motion is: \(m a = mg \sin \theta - 2kx\)
03

Simplify and rewrite as a linear differential equation

Now, let's make a substitution to simplify the equation: \(a = \frac{d^2x}{dt^2}\). This allows us to rewrite the equation of motion as a linear differential equation: \(m \frac{d^2x}{dt^2} = mg \sin \theta - 2kx\)
04

Find the harmonic constant

Rearranging the equation, we can find the harmonic constant associated with the oscillation: \(m\frac{d^2x}{dt^2}+ 2kx = mg \sin \theta\) The harmonic constant can be found by comparing this equation to the standard equation for simple harmonic motion: \(M\frac{d^2x}{dt^2}+ Kx = 0\) The harmonic constant, in this case, is \(K = 2k\).
05

Calculate the period of oscillation

To find the period of oscillation, we can use the relation between the period (T), the harmonic constant (K), and the mass (M): \[T = 2 \pi \sqrt{\frac{M}{K}}\] Here, \(M = m\) and \(K = 2k\). Substituting the values, we get: \[T = 2 \pi \sqrt{\frac{m}{2k}}\] Hence, the correct answer is (A) \(2 \pi \sqrt{\frac{m}{2k}}\).

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Most popular questions from this chapter

A pipe of length \(85 \mathrm{~cm}\) is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below \(1250 \mathrm{~Hz}\). The velocity of sound in air is \(340 \mathrm{~m} / \mathrm{s}\) (A) 12 . (B) 8 (C) 6 (D) 4

If \(Y_{1}=5 \sin (\omega t)\) and \(Y_{2}=5[\sqrt{3} \sin \omega t+\cos \omega t]\) are two SHMs, the ratio of their amplitude is (A) \(1: \sqrt{3}\) (B) \(1: 3\) (C) \(1: 2\). (D) \(1: \cos \left(\frac{\pi}{6}\right)\)

Two SHMs are represented by the equations: \(Y_{1}=10 \sin [3 \pi t+\pi / 4]\) \(Y_{2}=5[\sin 3 \pi t+\sqrt{3} \cos 3 \pi t]\) (A) The amplitude ratio of the two SHM is \(1: 1\). (B) The amplitude ratio of the two SHM is \(2: 1\). (C) Time periods of both the SHMs are equal. (D) Time periods of two SHMs are different.

An open glass tube is immersed in mercury in such a way that a length of \(8 \mathrm{~cm}\) extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional \(46 \mathrm{~cm}\). What will be length of the air column above mercury in the tube now? (Atmospheric pressure \(=76 \mathrm{~cm}\) of \(\mathrm{Hg}\) ) (A) \(16 \mathrm{~cm}\) (B) \(22 \mathrm{~cm}\) (C) \(38 \mathrm{~cm}\) (D) \(6 \mathrm{~cm}\)

A particle moves on the \(x\)-axis as per the equation \(x=x_{0} \sin ^{2} \omega t .\) The motion is simple harmonic (A) With amplitude \(x_{0}\) (B) With amplitude \(2 x_{0}\) (C) With time period \(\frac{2 \pi}{\omega}\) (D) With time period \(\frac{\pi}{\omega}\)
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