91Ó°ÊÓ

A light elastic spring, of natural length \(a\), and modulus \(8 m g\), stands vertically with its lower end fixed and carries a particle of mass \(m\) fastened to its upper end. This particle is resting in equilibrium when a second particle, also of mass \(m\), is dropped on to it from rest at a height \(3 a / 8\) above it. The particles coalesce on impact. Show that the composite particle oscillates about a point which is at a height \(\frac{3 a}{4}\) above the lower end of the spring and that the equation of motion is $$ \frac{d^{2} x}{d t^{2}}=-\frac{4 g x}{a} $$ where \(x\) is the displacement, at time \(t\), of the composite particle from its centre of oscillation. State the period and find the amplitude of the resulting motion. (Standard formulae for simple harmonic motion may be quoted without proof).

Step by step solution

01

- Analyze equilibrium position

Initially, the first particle is at equilibrium. Let this equilibrium position be at distance \(x_1\) from the natural length \(a\) of the spring. The force due to the weight of the particle \(mg\) is balanced by the restoring force of the spring, which is \(kx_1\). Modulus \(8mg\) implies \(k = 8mg/a\). Therefore, \(mg = kx_1\) results in \(mg = \frac{8mg}{a} x_1\), solving for \(x_1\) gives \(x_1 = \frac{a}{8}\).

02

- Impact calculation to find new equilibrium

The second particle of mass \(m\) is dropped from a height \(3a/8\). It will gain potential energy equal to \(\frac{3}{8}mg a\). On impact, the two particles coalesce forming a single particle of mass \(2m\). At the equilibrium position, \(kx_2 = 2mg\). Thus \(k = \frac{8mg}{a}\) and solving \(\frac{8mg}{a} x_2 = 2mg\), we get \(x_2 = \frac{a}{4}\).

03

- Calculate the center of oscillation

At the new equilibrium, the composite particle of mass \(2m\) oscillates about \(x_2 = a/4\) from the natural length of the spring. Therefore, the oscillation center is \(a + x_2 = \frac{3a}{4}\) from the lower end of the spring.

04

- Derive the equation of motion

Considering displacement \(x\) from the new equilibrium position and using Hooke's law, the restoring force is \(F = -kx\). The system follows Newton's second law: \(2m \frac{d^2 x}{d t^2} = -kx\). Given \(k = \frac{8mg}{a}\), this simplifies to \(\frac{d^2 x}{d t^2} = -\frac{4 g x}{a}\).

05

- Determine period of oscillation

The standard form for simple harmonic motion is \(\frac{d^2 x}{d t^2} = -\frac{k}{m} x\). Comparing with the derived equation \(\frac{d^2 x}{d t^2} = -\frac{4 g x}{a}\), it shows that \(\frac{k}{m} = \frac{4g}{a}\). The period \(T\) of simple harmonic motion is \(T = 2\frac{\frac{2\text{\tiny\bfseries )ve}}{\boldsymbol d\text<|vq_8573|>\frac{am}}}{\frac\frac}s}which simplifies to period to \)T_{\frac{\frac d\boldsymbol}\tfrac{k}{m}}\frac{\round} arc{c}\frac\(^\text{u}^{2\frac{k}{m}(m\frac{)\)arc).}

06

- Calculate the amplitude of motion

Initially, the second particle drops from a height \(\frac{3a}{8}\) and hits the first particle. This gives the amplitude of motion \(A = \frac{3a}{8}\) since this is the maximum displacement from equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Spring

An elastic spring is a spring that resists deformation and returns to its original shape once the force causing the deformation is removed. In this problem, we are dealing with a light elastic spring with a natural length denoted by \(a\) and modulus 8mg. The modulus, or stiffness constant \(k\), of the spring is given by k = \(\frac{8mg}{a}\). This means the spring can exert a force proportional to the displacement from equilibrium.

Equilibrium Position

The equilibrium position of a spring is the point where the force due to the spring's deformation balances the weight of the attached mass. Initially, a single particle of mass \(m\) is attached to the spring. In equilibrium, the force of gravity (mg) is balanced by the spring's restoring force (kx). Given k = \(\frac{8mg}{a}\), we find that x = \(\frac{a}{8}\). When a second mass of \(m\) is added, the new equilibrium position changes. Now, the total mass is 2m, and the new equilibrium position x' is \(\frac{a}{4}\). This position is from the spring's natural length.

Oscillation Period

The oscillation period is the time it takes for the particle to complete one full cycle of motion. For simple harmonic motion (SHM), the period T is given by \(T = 2\pi\sqrt{\frac{m}{k}}\). From the derived equation of motion \(\frac{d^2x}{dt^2} = -\frac{4gx}{a}\), we identify \(\frac{k}{m} = \frac{4g}{a}\). Therefore, \(T = 2\pi\sqrt{\frac{a}{4g}}\). This formula shows that the period depends only on the properties of the spring and the acceleration due to gravity.

Amplitude of Motion

The amplitude of motion in simple harmonic motion (SHM) is the maximum displacement from the equilibrium position. In this problem, the second particle drops from a height of \(\frac{3a}{8}\) and joins the first particle. Thus, just after impact, the maximum displacement, and hence the amplitude, will be \(\frac{3a}{8}\). This represents the farthest distance the composite particle will move from its new equilibrium position during oscillation.