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Displacement-time equation of a particle executing SHM is, \(x=4 \sin \omega t+3 \sin (\omega t+\pi / 3)\). Here \(x\) is in centimetre and \(t\) in second. The amplitude of oscillation of the particle is approximately (a) \(5 \mathrm{~cm}\) (b) \(6 \mathrm{~cm}\) (c) \(7 \mathrm{~cm}\) (d) \(9 \mathrm{~cm}\)

Short Answer

Expert verified
The amplitude is approximately 6 cm (option b).

Step by step solution

01

Identify the Form of the Oscillation Equation

The given displacement-time equation is \( x = 4 \sin \omega t + 3 \sin (\omega t + \pi / 3) \). This expression represents the combined effect of two simple harmonic motions.
02

Apply the Phase Addition Formula

We apply the trigonometric identity: \( \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \) to combine the two terms. This can help to find the resultant amplitude.
03

Decompose the Sine Terms

The first term has amplitude 4 and phase \( \phi_1 = 0 \). The second term has amplitude 3 and phase \( \phi_2 = \pi/3 \). Translate \( \sin(\omega t + \pi/3) \) into cosine terms using \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).
04

Combine the Sine Waves

Combine the two sine waves. Since amplitudes 4 and 3 add using the formula for the resultant amplitude of \( R = \sqrt{A^2 + B^2 + 2AB \cos \phi} \), where \( \phi \) is \( \pi/3 \).
05

Calculate the Resultant Amplitude

Substitute \( A = 4 \), \( B = 3 \), and \( \cos(\pi/3) = 0.5 \) into the formula: \[ R = \sqrt{4^2 + 3^2 + 2 \times 4 \times 3 \times 0.5} = \sqrt{16 + 9 + 12} = \sqrt{37} \approx 6.08 \text{ cm} \].
06

Determine the Closest Option

Since \( R \approx 6.08 \text{ cm} \), the amplitude of the oscillation of the particle is closest to option (b), which is \( 6 \text{ cm} \).

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Key Concepts

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

Amplitude Calculation
In simple harmonic motion (SHM), the amplitude is a critical parameter. It represents the maximum displacement from the equilibrium position.
In the given displacement-time equation:
\[ x = 4 \sin \omega t + 3 \sin (\omega t + \pi / 3) \]
You are dealing with the superposition of two harmonic motions.
Here's how we calculate the resulting amplitude:
  • The first component has amplitude 4.
  • The second component has amplitude 3.
To find the total amplitude when these oscillations are combined, use the formula:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \phi} \]
where:
  • \( A \) is 4 (amplitude of the first term)
  • \( B \) is 3 (amplitude of the second term)
  • \( \phi \) is the phase difference, which in this case is \( \pi / 3 \)
Substitute the given values:
\[ R = \sqrt{4^2 + 3^2 + 2 \times 4 \times 3 \times 0.5} = \sqrt{37} \approx 6.08 \text{ cm} \]
This shows that the resultant amplitude is approximately 6.08 cm, making it closest to 6 cm.
Trigonometric Identities
Trigonometric identities are essential tools in mathematics and physics, especially when dealing with oscillations and waves.
These identities simplify complex expressions and help in solving the displacement-time equation in SHM.
For this exercise, the identity for sum of sines is key:
  • \( \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \)
This identity allows the combination of two wave functions.
It results in a simplified form that can be easier to analyze or use in other calculations.
In this example, we used the identity to handle the two components \( 4 \sin \omega t \) and \( 3 \sin (\omega t + \pi / 3) \).
By knowing such identities, we turn tricky trigonometric problems into more manageable parts!
Phase Addition Formula
Phase addition formulas are used to combine wave functions with different phases. They are quite handy in SHM.
For any two sine terms, like in this problem:
  • First term: \( 4 \sin \omega t \) with no additional phase
  • Second term: \( 3 \sin (\omega t + \pi / 3) \) with a phase shift
To find the resulting phase and amplitude, the phase addition formula comes into play:
The standard approach involves the identity:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
This breaks down the terms for easier addition or manipulation.
Given that both components have specific amplitudes, their phase difference \( \phi = \pi / 3 \) impacts the overall oscillation pattern.
Understanding phase relationships in waves enables us to calculate superpositions like this one accurately.
It's just another example of how many different aspects of wave physics are interconnected!

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Most popular questions from this chapter

The displacement of a particle is represented by the equation \(y=3 \cos \left(\frac{\pi}{4}-2 \omega t\right)\). The motion of the particle is [NCERT Exemplar] (a) simple harmonic with period \(2 \mathrm{p} / \omega\) (b) simple harmonic with period \(\pi / \omega\) (c) periodic but not simple harmonic (d) non-periodic

When a body of mass \(1.0 \mathrm{~kg}\) is suspended from a certain light spring hanging vertically, its length increases by \(5 \mathrm{~cm}\). By suspending \(2.0 \mathrm{~kg}\) block to the spring and if the block is pulled through \(10 \mathrm{~cm}\) and released, the maximum velocity of it, in \(\mathrm{ms}^{-1}\) is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(0.5\) (b) 1 (c) 2 (d) 4

A point particle of mass \(0.1 \mathrm{~kg}\) is executing SHM of amplitude \(0.1 \mathrm{~m}\). When the particle passes through the mean position, its kinetic energy is \(8 \times 10^{-3} \mathrm{~J}\). The equation of motion of this particle, if its initial phase of oscillation is \(45^{\circ}\), is (a) \(y=0.1 \sin \left(\frac{r}{4}+\frac{\pi}{4}\right) \quad\) (b) \(y=0.1 \sin \left(\frac{t}{2}+\frac{\pi}{4}\right)\) (c) \(y=0.1 \sin \left(4 t-\frac{\pi}{4}\right) \quad\) (d) \(y=0.1 \sin \left(4 t+\frac{\pi}{4}\right)\)

A particle of mass \(m\) is executing oscillations about the origin on the \(x\)-axis with amplitude \(A\). Its potential energy is given as \(U(x)=\alpha x^{4}\), where \(\alpha\) is positive constant. The \(x\)-coordinate of mass where potential energy is one-third of the kinetic energy of particle, is (a) \(\pm \frac{A}{\sqrt{3}}\) (b) \(\pm \frac{A}{\sqrt{2}}\) (c) \(\pm \frac{A}{3}\) (d) \(\pm \frac{A}{2}\)

Assertion The amplitude of a particle executing SHM with a frequency of \(60 \mathrm{~Hz}\) is \(0.01 \mathrm{~m}\). The maximum value of acceleration of the particle is \(\pm 144 \pi^{2} \mathrm{~ms}^{-2}\). Reason Acceleration amplitude \(=\omega^{2} A\),where \(A\) is displacement amplitude.

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