/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 157 Two identical straight wires are... [FREE SOLUTION] | 91Ó°ÊÓ

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Two identical straight wires are stretched so as to produce 6 beats per sec when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency remains unchanged. Denoting by \(T_{1}, T_{z}\), the higher and the lower initial tensions in the strings, then it could be said that while making the above changes in tension (a) \(T_{2}\) was decreased (b) \(T_{1}\) was increased (c) \(T_{2}\) was increased (d) \(T_{1}\) was decreased

Short Answer

Expert verified
Option (c): \(T_{2}\) was increased.

Step by step solution

01

Understanding Beat Frequency

When two frequencies are close to each other, they produce a beat frequency, which is the absolute difference between the two frequencies. In this case, 6 beats per second indicate that the difference between the frequencies is 6 Hz.
02

Effect of Tension on Frequency

The frequency of a vibrating wire is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length, \( T \) is the tension, and \( \mu \) is the mass per unit length. Increasing tension increases frequency.
03

Initial Setup with Equal Tensions

Initially, both wires should be at similar tensions but slightly different, leading to the 6 Hz beat frequency due to a small difference in their frequencies.
04

Change in Tension and its Impact

When one of the strings' tension is slightly changed, if the beat frequency remains unchanged at 6 beats per second, the new difference between tensions should still result in frequencies differing by 6 Hz.
05

Analyzing the Options

(a) Decreasing \( T_2 \) (the lower tension) would further lower its frequency, increasing the beat frequency. (b) Increasing \( T_1 \) (the higher tension) would increase its frequency, affecting the beat frequency. (c) Increasing \( T_2 \) brings its frequency closer to \( T_1 \), keeping the 6 Hz difference. (d) Decreasing \( T_1 \) would again alter the beat frequency beyond 6 Hz.
06

Conclusion

From analyzing the conditions, increasing \( T_2 \) is the change that maintains the beat frequency at 6 Hz.

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Key Concepts

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

Vibrating Wires
In the context of sound waves and vibrations, wires or strings can vibrate to produce audible sound frequencies. These frequencies are determined by the physical properties of the wires, such as length and mass per unit length. When wires vibrate, they generate sound through their natural resonant frequencies. These vibrating wires act in similar ways to musical instruments, creating musical notes based on their vibrational modes. When two wires are placed close to each other and play at near similar frequencies, they can produce a fascinating phenomenon called beats. This happens when their sound waves interfere constructively and destructively at various points in time, leading to alternating loud and soft sounds that we perceive as beats. The rate of these beats is termed the "beat frequency". This concept is crucial in understanding how small changes in the tension of a wire can impact overall sound frequency.
Tension and Frequency
The relationship between tension and frequency is key to understanding how stringed instruments work, as well as many physics problems. The tension in a wire directly influences its frequency of vibration. The mathematical representation of this relationship is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]Where:
  • \( f \) is the frequency of the vibration
  • \( L \) is the length of the wire
  • \( T \) is the tension in the wire
  • \( \mu \) is the mass per unit length of the wire
This formula shows that increasing the tension in the wire (\( T \)) will result in an increased frequency. Thus, if you slightly change the tension in one wire while comparing it with another, it can alter the resulting beat frequency. This is why understanding how to control tension is vital when trying to match or manipulate the frequencies of two wires to maintain or alter their beat frequency.
Harmonic Motion
Harmonic motion is a fundamental concept in physics that describes systems that return to equilibrium positions through regular, periodic oscillations. This is often described as a sine wave in physics, representing continuous and smooth oscillation. In the case of vibrating wires, each wire is undergoing some form of harmonic motion. As the wire vibrates, the displacement versus time graph of the vibration typically takes the form of a sinusoidal wave. The studying of such harmonic oscillations is important because it provides insight into how energy is transferred within the wire, and how those oscillations merge or cancel each other out to produce the beat frequency when in proximity to another wire. It's useful for students to visualize how harmonic motion underlies many natural phenomena, providing patterns for waves, including sound, light, and electromagnetic waves, which operate on similar principles even if the medium and forces involved vary.
JEE Physics Problem
JEE (Joint Entrance Examination) physics problems often incorporate real-world applications of theoretical concepts. The exercise about vibrating wires and beat frequency serves as a great example of this. Students preparing for exams such as the JEE need to grasp how to apply formulas and physical understanding to solve complex problems. In the exam context, students must decipher which physical principles apply, such as the relationship between tension and frequency when adjusting wire tension to yield a particular beat frequency. Such problems challenge students not only to plug numbers into formulas but to critically think about how physical properties interact. These scenarios are designed to enhance the student's ability to think through logical connections between concepts like tension, frequency, and beat phenomena. Developing skills in these areas is crucial for those who wish to pursue further studies and careers in fields requiring advanced understanding of physics.

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Most popular questions from this chapter

(a) Let \(L \rightarrow\) stationary, \(\nu=580 \mathrm{II} \mathrm{z}\) \(s \rightarrow\) moving with \(40 \mathrm{~km} / \mathrm{hr}\) and aproaching \(w \rightarrow 40 \mathrm{~km} / \mathrm{hr}\) and supporting \(v=1200 \mathrm{~km} / \mathrm{hr}\) So \(v^{\prime}=\left[\frac{v+\omega}{(v+\omega)-v_{s}}\right] v\) \(\Rightarrow \quad v^{\prime}=\frac{1240}{1200} \times 580=\frac{1798}{3}=599.33 \mathrm{IIz}\) (b) At a distance \(1 \mathrm{~km}\) before the train whistles and let the driver heard the echo at time \(l=t^{\prime}\). then at this time, total distance travelled by wave and train \(=2 \mathrm{~km}\) Now time after which first wave reach to the hill in time interval \(t_{1}=\frac{1}{1200} \mathrm{hr}\) after reflecting at \(t=t^{\prime}\), echo will heard, then distance travellcd by it is \(x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200\) and distance travelled by the train is \(x_{2}=\left(t^{\prime}-\frac{1}{1200}\right) 40\) As \(x_{1}+x_{2}=1 \mathrm{~km}\) \(\therefore\left(t^{\prime}-\frac{1}{1200}\right) \times 12040+\left(t^{\prime}-\frac{1}{1200}\right) 40\) \(1240 t^{\prime}=\frac{1}{30} \quad \therefore \quad t^{\prime}=\frac{61}{30 \times 1240}\) and required distance, $$ x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200 $$$=\left(\frac{61}{30 \times 1240}-\frac{1}{1200}\right) \times 1200=\frac{61 \times 1200}{30 \times 1240}-1\( \)=\frac{30}{31} \times 1000 \mathrm{~m}=967.74 \mathrm{~m}\( Frequency of sound heard by the driver, \)v^{\prime \prime}=\left(\frac{(v-\omega)+v_{2}}{v-\omega}\right) v\( \)=\left(\frac{1200-40+40}{1200-40}\right) \times 599.33\( \)\therefore \quad v^{\prime \prime}=\frac{1200}{1160} \times 599.33=620 \mathrm{~Hz}$

Statement-1 : \(\Lambda\) particle executes SIIM with frequency \(f .\) The frequency with which total mechanical energy of particle oscillates is \(2 f\). Statement-2 : Both kinetic as well as potential energy of particle in SIIM oscillate with a frequency \(2 f\).

As \(T=2 \pi \sqrt{\frac{I}{c}}\) where \(I=\) M.I. of the system, \(c=\) torsional constant \(\therefore \quad 2=2 \pi \sqrt{\frac{0.2}{c}} \Rightarrow \frac{2 \pi}{\sqrt{c}}=\frac{2}{\sqrt{0.2}}\) Now when new disc is loaded with another disc, then \(T=2 \cdot 5=2 \pi \sqrt{\frac{I^{\prime}}{c}}=\frac{2 \pi}{\sqrt{c}} \sqrt{l^{\prime}}\) \(\Rightarrow \quad 2.5=\frac{2}{\sqrt{0.2}}\), from (1) \(\Rightarrow \quad 6.25=\frac{4}{0.2} \times l^{\prime} \quad \therefore \quad l^{\prime}=0.315\) Now required moment of inertia \(=0.315-0.2=0.115 \mathrm{~kg}-\mathrm{m}^{2}\)

\(\delta=\frac{m}{L}=\frac{32}{4} \times 10^{2}=\frac{4}{5} \times 10^{2} \mathrm{~kg} / \mathrm{m}\) Now, According to question, \(v=v_{o}=220 \mathrm{II} z\) \(\therefore \frac{v}{2 L}=220 \mathrm{~Hz} \quad[\quad L=40 \mathrm{~cm}\) \(m=3.2 \mathrm{gm}\) \(\sqrt{\frac{T}{\delta}} \times \frac{1}{2 L}=220 \mathrm{II} z\) 1. \(\sqrt{\frac{T}{\rho}}=220 \times 2 L\) \(\therefore \quad \sqrt{T}=\sqrt{\rho} \times 220 \times 2 L$$$ \begin{aligned} &\therefore \quad T=\rho \times(220 \times 2 L)^{2}=\frac{4}{5} \times 10^{-2}(22 \times 2 \times 4)^{2} \\ &=\frac{4}{5} \times 10^{-2} \times 30976=2478.8 \times 10^{-2} \\ &\text { Now suress }=\frac{F}{\Lambda}=\frac{T}{\Lambda}=\frac{2478.8 \times 10^{-2}}{\left(10^{-3}\right)^{2}} \\ &=2478.8 \times 10^{4}=\mathrm{N} / \mathrm{m}^{2} \end{aligned} $$ and strain $$ =\frac{0.05}{40} \times \frac{10^{2}}{10^{2}}=\frac{5}{40} \times 10^{-2}=\frac{1}{8} \times 10^{-2} $$ \)\therefore$ (Youngs modulus) $$ \begin{aligned} Y &=\frac{\text { stress }}{\text { strain }}=\frac{2478.8 \times 10^{4}}{10^{2} / 8} \\ &=1.98 \times 101^{\prime} \mathrm{N} / \mathrm{m}^{2} \end{aligned} $$

Which one of the following represent a travelling wave? (a) \(y-A \sqrt{(x-v)}\) (b) \(y=A \cos (a x+b c)\) (c) \(y=A \log (x-v t)\) (d) \(y=f\left(x^{2}-y^{2}\right)\)

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