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Two waves are represented by \(\mathrm{y}_{1}=\mathrm{A} \sin \omega \mathrm{t}\) and \(\mathrm{y}_{2}=\mathrm{A}\) cos \(\omega \mathrm{t}\). The phase of the first wave, \(\mathrm{w}\). r. t. to the second wave is (A) more by radian (B) less by \(\pi\) radian (C) more by \(\pi / 2\) (D) less by \(\pi / 2\)

Short Answer

Expert verified
The phase of the first wave relative to the second wave is less by \(\frac{\pi}{2}\) radians. Hence, the correct answer is (D).

Step by step solution

01

Recall the relationship between sine and cosine functions

The sine and cosine functions are related by the following identity: \[\sin(x) = \cos\left( x - \frac{\pi}{2} \right)\] This means that the sine function is simply a shifted version of the cosine function by a phase of \(\frac{\pi}{2}\).
02

Compare the given waves

We have the two waves, \(y_1 = A\sin(\omega t)\) and \(y_2 = A\cos(\omega t)\). We can rewrite \(y_1\) using the identity mentioned in Step 1: \[y_1 = A\sin(\omega t) = A\cos\left(\omega t - \frac{\pi}{2}\right)\]
03

Determine the phase difference

Comparing the rewritten equation of \(y_1\) with that of \(y_2\), we can see that the first wave has a phase that is less than the second wave by \(\frac{\pi}{2}\) radians. Therefore, the correct answer is (D) less by \(\frac{\pi}{2}\)

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Key Concepts

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

Trigonometric identities
Trigonometric identities are essential tools in mathematics that help us understand relationships between different trigonometric functions, such as sine, cosine, and tangent. These identities allow us to rewrite and manipulate expressions involving trigonometric functions to solve problems more easily.
For instance, the identity \[sin(x) = \cos\left(x - \frac{\pi}{2}\right) \]demonstrates how the sine function is related to the cosine function. Essentially, this identity tells us that the sine of an angle is equivalent to the cosine of the angle after a phase shift of \(\frac{\pi}{2}\) radians. This phase shift is critical because it helps us understand the timing and synchronization between these functions, especially in contexts like wave interference.
Applying these identities helps simplify complex mathematical problems and provides a clearer interpretation of how different waves interact with each other.
Wave interference
Wave interference occurs when two or more waves overlap and combine to form a new wave pattern. The principle of superposition is key to understanding how this happens, as it states that the resulting wave is the sum of the individual waves.
Wave interference can be constructive or destructive, depending on the phase relationship between the overlapping waves.
  • Constructive interference: Happens when waves align in such a way that their amplitudes add together, resulting in a wave of greater amplitude.
  • Destructive interference: Occurs when waves are out of phase, and their amplitudes subtract from each other, causing a reduction in overall wave amplitude or even cancellation.
Consider the given waves from the original exercise, where both are defined by sine and cosine functions. By understanding the phase difference, we can predict how these waves will interact. In this exercise, a phase shift of \(\frac{\pi}{2}\) radians is applied, showing that even small phase differences can greatly impact the resulting wave patterns.
Phase shift
A phase shift refers to the horizontal displacement between two identical waveforms or a change in their starting points. It's crucial in the study of wave behavior because it can modify the way two waves interact when combined.
In mathematical expressions like those in the original exercise, a phase shift can be seen in the comparison between sine and cosine functions. Initially, \[y_1 = A\sin(\omega t) \]can be rewritten using the trigonometric identity \[y_1 = A\cos\left(\omega t - \frac{\pi}{2}\right)\]This indicates a phase shift of \(\frac{\pi}{2}\) radians between the two functions.
Phase shifts are significant because they determine how wave peaks and troughs line up. In practical applications, such as noise-canceling headphones or radio signal transmission, a correct understanding of phase shifts ensures that waves are manipulated accurately for desired outcomes.

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Most popular questions from this chapter

Two simple pendulums having lengths \(144 \mathrm{~cm}\) and \(121 \mathrm{~cm}\) starts executing oscillations. At some time, both bobs of the pendulum are at the equilibrium positions and in same phase. After how many oscillations of the shorter pendulum will both the bob's pass through the equilibrium position and will have same phase ? (A) 11 (B) 12 (C) 21 (D) 20

For a particle executing \(\mathrm{S} . \mathrm{H} \mathrm{M} .\), when the potential energy of the oscillator becomes \(1 / 8\) the maximum potential energy, the displacement of the oscillator in terms of amplitude A will be........... (A) \((\mathrm{A} / \sqrt{2})\) (B) \(\\{\mathrm{A} /(2 \sqrt{2})\\}\) (C) \((\mathrm{A} / 2)\) (D) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a \(2 \mathrm{~kg}\) mass in it. When the spring (massless) is pressed slightly and released, the \(2 \mathrm{~kg}\) mass, starts executing S.H.M. The force constant of the spring is \(200 \mathrm{Nm}^{-1}\). For what minimum value of amplitude, will the mass loose contact with the plate? (Take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(10.0 \mathrm{~cm}\) (B) \(8.0 \mathrm{~cm}\) (C) \(4.0 \mathrm{~cm}\) (D) For any value less than \(12.0 \mathrm{~cm}\).

One end of a mass less spring having force constant \(\mathrm{k}\) and length \(50 \mathrm{~cm}\) is attached at the upper end of a plane inclined at an angle \(e=30^{\circ} .\) When a body of mass \(m=1.5 \mathrm{~kg}\) is attached at the lower end of the spring, the length of the spring increases by \(2.5 \mathrm{~cm}\). Now, if the mass is displaced by a small amount and released, the amplitude of the resultant oscillation is ......... (A) \((\pi / 7)\) (B) \((2 \pi / 7)\) (C) \((\pi / 5)\) (D) \((2 \pi / 5)\)

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of \((\mathrm{g} / 3)\), then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)

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