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When two waves of almost equal frequencies \(n_{1}\) and \(n_{2}\) are produced simultaneously, then the time interval between successive maxima is (a) \(\frac{1}{n_{1}-n_{2}}\) (b) \(\frac{1}{n_{1}}-\frac{1}{n_{2}}\) (c) \(\frac{1}{n_{1}}+\frac{1}{n_{2}}\) (d) \(\frac{1}{n_{1}+n_{2}}\)

Short Answer

Expert verified
The time interval is (a) \(\frac{1}{n_{1}-n_{2}}\).

Step by step solution

01

Understand the Concept of Beats

When two waves of slightly different frequencies interfere, they produce a phenomenon called 'beats.' The beat frequency is the absolute difference between the two frequencies, giving rise to fluctuations in amplitude.
02

Mathematical Representation of Beat Frequency

The beat frequency can be calculated as the difference between the two frequencies. It is given by the formula: \( f_{beat} = |n_{1} - n_{2}| \). This frequency tells us how often the beats (or the amplitude maxima) occur.
03

Time Interval Between Successive Maxima

The time interval between successive maxima (or one beat) is the reciprocal of the beat frequency. Thus, the time interval \( T_{beat} \) between successive maxima is given by: \( T_{beat} = \frac{1}{f_{beat}} \).
04

Substitute Beat Frequency into Time Interval Formula

Now substitute the expression for beat frequency into the formula for the time interval: \( T_{beat} = \frac{1}{|n_{1} - n_{2}|} \). Since the difference in frequencies is a positive number, the formula becomes \( T_{beat} = \frac{1}{n_{1} - n_{2}} \).

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Key Concepts

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

Time Interval Between Maxima
When two waves with nearly equal frequencies interact, they create an intriguing rhythmic pattern known as 'beats.' These beats result from interference, where the wave amplitudes fluctuate in a predictable manner. The prime question here is: how do we determine the time interval between these peaks, or maxima, in the wave interference pattern?
To solve this, it's essential to grasp the concept of beat frequency. The beat frequency is defined as the absolute difference between the two frequencies of the interacting waves. So, if you have two waves with frequencies of \( n_{1} \) and \( n_{2} \), the beat frequency, \( f_{beat} \), is \( |n_{1} - n_{2}| \).
Once you have the beat frequency, calculating the time interval between successive maxima becomes straightforward. The reciprocal of the beat frequency gives the time interval \( T_{beat} \). Mathematically, this is expressed as:
  • \( T_{beat} = \frac{1}{f_{beat}} \)
Thus, the larger the frequency difference (i.e., the higher the beat frequency), the shorter the time between peaks. Ultimately, you end up with:
  • \( T_{beat} = \frac{1}{|n_{1} - n_{2}|} \)
This formula provides a clear path forward for finding the time between amplitude maxima when dealing with wave interference caused by similar frequencies.
Interference of Waves
Interference of waves happens when two or more waves traverse the same space, leading to a new wave pattern. This interaction can be constructive or destructive based on the phase relationship of the contributing waves.
When waves interfere constructively, they achieve a maximum when the peaks (crests) of the waves from different sources coincide. Conversely, destructive interference occurs when the peaks align with troughs, resulting in cancellation and minimal amplitude.

Role of Phase in Interference

Phase differences play a crucial role in determining if interference will be constructive or destructive. If two waves are "in phase," meaning their crests and troughs align perfectly, you'll observe constructive interference with amplified peaks.
If they are "out of phase" by half a wavelength, the waves tend to cancel each other out through destructive interference.
Understanding these concepts is crucial when analyzing the interference patterns, such as those seen in beats. While the two waves may differ only slightly in frequency, their superposition creates a rhythmic oscillation in amplitude that repeats over time.
Wave Frequency Difference
Understanding the difference between two wave frequencies is vital, especially when discussing phenomena like beats. The frequency difference, or \( \Delta f \), is the absolute numerical difference between two frequencies. For waves with frequencies \( n_{1} \) and \( n_{2} \), the frequency difference is expressed as:
  • \( \Delta f = |n_{1} - n_{2}| \)
Even a tiny difference can lead to noticeable effects, such as audible beats in sound waves. The closer the two frequencies, the lower the beat frequency, leading to slower oscillations in amplitude.

Why Frequency Difference Matters

Although the original frequencies of the waves might be high, the frequency difference can be small enough to be perceived more easily by human senses. This is especially true in audio applications, where beats can be detected as a warbling effect or a rhythm in a tone.
Also, in practical applications, measuring the frequency difference can help in tuning musical instruments or in scientific experiments where precise frequency control is necessary.
A firm grip on the concept of frequency differences enables the prediction and analysis of the interference patterns and is key in computing the time interval between amplitude maxima.

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Most popular questions from this chapter

The equation of a stationary wave along a stretched string is given by \(y=4 \sin \frac{2 \pi x}{2} \cos 40 \pi t\) where \(x\) and \(y\) are in \(\mathrm{cms}\) and \(t\) is in sec. The separation between two adjacent nodes is (a) \(3 \mathrm{~cm}\) (b) \(1.5 \mathrm{~cm}\) (c) \(6 \mathrm{~cm}\) (d) \(4 \mathrm{~cm}\)

When 2 tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats \(\mathrm{s}^{-1}\) are heard. Now, some tape is attached on the prong of fork \(2 .\) When the tuning forks are sounded again, 6 beats \(\mathrm{s}^{-1}\) are heard if the frequency of fork 1 is \(200 \mathrm{~Hz}\), then what was the original frequency of fork 2 ? (a) \(196 \mathrm{~Hz}\) (b) \(200 \mathrm{~Hz}\) (c) \(202 \mathrm{~Hz}\) (d) \(204 \mathrm{~Hz}\)

The ends of a stretched wire of length \(L\) are fixed at \(X=0\) and \(X=L\). In one experiment the displacement of the wire is \(Y_{1}=A \sin \left(\frac{\pi x}{L}\right) \sin \omega t\) and energy is \(E_{1}\) and in another experiment its displacement is \(y_{2}=A \sin (2 \pi x / L) \sin 2 \omega t\) and energy is \(E_{2}\). Then (a) \(E_{2}=E_{1}\) (b) \(E_{2}=2 E_{1}\) (c) \(E_{2}=4 E_{1}\) (d) \(E_{2}=16 E_{1}\)

\- A string of mass \(0.2 \mathrm{~kg} / \mathrm{m}\) has length \(l=0.6 \mathrm{~m}\). It is fixed at both ends and stretched such that it has a tension of \(80 \mathrm{~N}\). The string vibrates in three segments with amplitude \(=0.5 \mathrm{~cm}\). The amplitude of transverse velocity is (a) \(9.42 \mathrm{~ms}^{-1}\) (b) \(3.14 \mathrm{~ms}^{-1}\) (c) \(1.57 \mathrm{~ms}^{-1}\) (d) \(6.28 \mathrm{~ms}^{-1}\)

A source of sound emitting a tone of frequency \(200 \mathrm{~Hz}\) moves towards an observer with a velocity \(v\) equal to the velocity of sound. If the observer also moves away from the source with the same velocity \(v\), the apparent frequency heard by the observer is (a) \(50 \mathrm{~Hz}\) (b) \(100 \mathrm{~Hz}\) (c) \(150 \mathrm{~Hz}\) (d) \(200 \mathrm{~Hz}\)

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