Waves

When a rod is clamped at its midpoint, the midpoint must be a node (a point of zero displacement).
The ends of the rod are free to vibrate, so they must be antinodes (points of maximum displacement).
For the fundamental frequency (the simplest mode of vibration), the pattern will be Antinode - Node - Antinode.
The distance between an antinode and the next node is always one-quarter of a wavelength ( ).
The total length of the rod, L, covers the distance from one antinode end to the central node, plus the distance from the central node to the other antinode end.
So, .
This means the wavelength of the fundamental mode is .
We are given the length of the rod, .
So, the fundamental wavelength is m.
The relationship between frequency (f), wavelength ( ), and the speed of the wave (v) is .
We can find the fundamental frequency as .
We are given the speed of sound in the metal, m/s.
Hz.
The question asks for the answer in kHz.
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For a tube closed at one end, the possible wavelengths are given by , where L is the length of the tube and n is the harmonic number (n=1, 3, 5,... for closed pipes). The question says "fifth harmonic", which corresponds to the harmonic number n=3 in the series of possible harmonics for a closed pipe (1st, 3rd, 5th...). Let's assume "fifth harmonic" means . This would be the 9th overtone which is unusual. Let's assume it means the 5th possible resonant mode.
Let's take "fifth harmonic" to mean the mode with 5 quarter-wavelengths. For a closed pipe, the nth harmonic has a frequency of . The 5th harmonic would be for n=3. Let's assume the question means n=5. Then the mode is the 9th overtone. This is likely a misstatement, and it means the 2nd overtone, which is the 5th harmonic for an open pipe. Let's assume it's the 5th natural frequency for a closed pipe, which is in the formula . Wait, the harmonics are named by the multiple of the fundamental. So 5th harmonic is . A closed pipe only has odd harmonics. So this is the 5th harmonic.
The length L contains quarter wavelengths. So for the 5th harmonic, is not right. For the nth harmonic . For a closed pipe, only odd n are possible. So the 5th harmonic is when n=5.
So for the 5th harmonic, .
Given cm = 0.5 m.
m = 40 cm.
The phase difference between two points separated by a distance is given by .
One point is at the open end ( ). The other point is at cm from the open end.
The distance between them is cm.
Let's calculate the phase difference.
.
This isn't a simple angle. Let's re-read the question. A standing wave is formed.
In a standing wave, all particles between two consecutive nodes vibrate in phase. And particles in adjacent segments (separated by a node) vibrate in opposite phase ( difference).
The open end is an antinode. The closed end is a node. The positions of the nodes from the open end are at .
With cm, the nodes are at:
cm.
cm.
cm (at the closed end).
The antinodes are at , which are 0 cm, 20 cm, 40 cm.
The points are at the open end ( ) and at cm.
The segment between node at 30cm and node at 50cm contains the point cm. The antinode for this segment is at 40cm.
The point at the open end (x=0) is in the first segment (between x=0 and node at x=10). The point at x=42cm is in the third segment (between node at x=30 and node at x=50).
The phase relationship is: Segment 1 (0-10cm): phase Segment 2 (10-30cm): phase Segment 3 (30-50cm): phase , which is the same as .
Since the point at x=0 (open end) and the point at x=42 cm are in segments with the same phase, the phase difference between them is 0 or .
Therefore, the phase difference is .

Step 1: Concept (Fresnel Distance):
The distance up to which ray optics is a good approximation is called the Fresnel distance ( ). Formula: where is the aperture size and is the wavelength.
Step 2: Calculation:
Given , .
Step 3: Convert to Angstroms:
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Step 1: Formulas for Harmonics:
For an Open Pipe of length : Fundamental frequency . harmonic frequency . Third harmonic ( ): . For a Closed Pipe of length : Fundamental frequency . Only odd harmonics exist: . Fifth harmonic corresponds to frequency : .
Step 2: Calculate Ratio:
Given ratio of lengths . Ratio of frequencies: Final Answer:
The ratio is .

Step 1: Understanding the Concept:
The formation of coloured images (defects) by lenses is called Chromatic Aberration. This occurs because the focal length of a lens depends on the refractive index ( ). Since the refractive index varies with wavelength (color) of light, different colors focus at different points.
Step 2: Identifying the Property:
The phenomenon where refractive index depends on wavelength, causing splitting of white light into constituent colors, is called Dispersion.

When two sound waves of slightly different frequencies interfere, they produce a phenomenon called beats.
The beat frequency ( ) is the difference between the two frequencies.
Hz.
Hz.
Hz.
The beat frequency represents the number of times the sound intensity becomes maximum per second. So, there are 3 maxima (or "beats") per second.
The time interval between two consecutive maxima is the beat period, .
s.
The sound intensity varies sinusoidally between maximum and minimum. A full cycle consists of a maximum followed by a minimum, then another maximum.
The time interval between a maximum and the very next (adjacent) minimum is half of the beat period.
Time interval = s.

This is a problem involving the Doppler effect for sound.
Let be the actual frequency of the stationary source.
Let be the speed of sound in air, m/s.
Let be the speed of the observer.
Case 1: Observer moving towards the stationary source.
The apparent frequency heard is .
Case 2: Observer moving away from the stationary source.
The apparent frequency heard is .
We are given the ratio .
.
So, .
Cross-multiply to solve for :
.
.
.
.
.
Substitute the value of m/s:
m/s.
The options are in kmph. We need to convert the speed.
To convert m/s to kmph, multiply by .
Speed of observer = kmph.

Step 1: Understanding the Concept:

The question asks for the standard bandwidth allocated for voice transmission in commercial telephone systems. It is important to distinguish between the full range of human hearing and the specific range required for intelligible speech.

Step 2: Analyzing Frequency Ranges:

• Human Audible Range:
  The human ear can detect frequencies from approximately to . This range is necessary for high-fidelity audio (like music) to sound natural and full.
  
• Speech Spectrum:
  Although human speech contains frequencies extending up to or more, the majority of the energy and the components necessary for intelligibility (understanding words and recognizing the speaker) are concentrated in the lower frequency range.
  
• Telephony Standards:
  Commercial telephone systems limit the bandwidth to save transmission capacity. The standard "voice band" defined for telephony allows frequencies roughly between and to pass. This range captures the fundamental frequency of the voice and the significant formants required for speech recognition.
  

Step 3: Evaluating the Options:

• Option (1) 20 Hz - 20 kHz:
  This is the entire audible range. Transmitting this wide band is unnecessary for voice communication and inefficient for commercial telephony.
  
• Option (2) 300 Hz - 3100 Hz:
  This range falls directly within the standard voice band ( to ). It is the accepted range for commercial telephonic communication to ensure speech is intelligible while conserving bandwidth.
  
• Option (3) 200 MHz - 600 MHz:
  This is in the Ultra High Frequency (UHF) range, typically used for television broadcasting and cellular carrier frequencies, not baseband voice signals.
  
• Option (4) 300 kHz - 8000 kHz:
  This falls into the RF spectrum (Medium and High Frequency) used for radio broadcasting (AM/Shortwave), not for the audio signal itself.
  

Step 4: Final Conclusion:

The adequate frequency range for speech signals in commercial telephony is the voice band, which corresponds to option (2).

Step 1: Understanding the Configuration:
Objective focal length . Eyepiece focal length . Separation distance (often approximated as tube length in competitive exams) . Final image is at infinity (Normal adjustment).
Step 2: Formula for Magnification:
When the final image is at infinity, the magnification is: where is the least distance of distinct vision ( ) and is the distance between the lenses (or tube length approximation used here).
Step 3: Calculation:
Note: While strictly in the formula is the distance between foci, in many exam contexts, the separation distance given is used directly as for the simplified formula.

Step 1: Understanding the Concept:
The fundamental frequency ( ) of a vibrating string is determined by its length ( ), the tension ( ) applied to it, and its linear mass density ( ). We need to analyze how changes in length and tension affect this frequency.
Step 2: Key Formula or Approach:
The formula for the fundamental frequency of a string is:
We can set up a ratio of the final frequency ( ) to the initial frequency ( ) to solve for the ratio of tensions .
Step 3: Detailed Explanation:
Let the initial conditions be and the final conditions be . The linear density is constant.
Initial frequency: .
Now, let's define the final conditions based on the problem statement:
- "length of the string is decreased by 25%": .
- "fundamental frequency... increases by 100%": .
Final frequency: .
Now, we take the ratio of the frequencies:
The terms and cancel out.
Substitute the known values for the ratios of frequency and length:
Now, we solve for the ratio of tensions. First, isolate the square root term:
Finally, square both sides to get the desired ratio:
Step 4: Final Answer:
The ratio is . Therefore, option (D) is correct.

Step 1: Understanding the Concept:
The equation for a one-dimensional sinusoidal wave traveling in the positive x-direction is given by . We need to determine the amplitude (A), the wave number (k), and the angular frequency ( ) from the given information.
Step 2: Key Formula or Approach:
- Amplitude ( ): The maximum displacement from the equilibrium position.
- Angular frequency ( ): , where is the frequency.
- Wave number ( ): , where is the wavelength.
- Wave speed ( ): .
Step 3: Detailed Explanation:
Determine the Amplitude (A):
The problem states that "each particle of the wave moves a distance of 10 cm between the two extreme points". The distance between the two extreme points of an oscillation is twice the amplitude ( ).
This immediately eliminates options (A) and (D).
Determine the Angular Frequency ( ):
Given frequency Hz.
Looking at the options, is not present. This suggests an approximation. Let's calculate the numerical value. rad/s. This is very close to 1320 rad/s. Let's assume rad/s.
Determine the Wave Number (k):
We are given the wave speed m/s. We can use the relation .
Construct the Wave Equation:
Now we assemble the equation .
This matches option (B). The form is because the wave travels along the positive x-axis. Option (C) has the values for k and swapped, which is incorrect.
Step 4: Final Answer:
The correct equation for the displacement function is m. Therefore, option (B) is correct.

Step 1: Understanding the Concept:

When two coherent waves interfere, the resultant intensity depends on their individual intensities and the phase difference between them.
Step 2: Key Formula or Approach:

Resultant Intensity Formula:
Step 3: Detailed Explanation:

Given: , Phase difference Substitute into the formula: Since :
Step 4: Final Answer:

The resultant intensity is 2I.

Step 1: Understanding the Concept:

The fundamental frequency of a stretched string depends on its length, tension, and linear mass density.
Step 2: Key Formula or Approach:

Formula for fundamental frequency: where is length, is tension, and is linear mass density.
Step 3: Detailed Explanation:

Given: Hz, , Hz, , is constant. Write the ratio : Squaring both sides:
Step 4: Final Answer:

The ratio is 4:9.

• One full cycle contains one crest and one trough. The presence of 60 crests and 60 troughs in 0.2 s corresponds to 60 full cycles in 0.2 s.
• Frequency .
• Distance between crest and adjacent trough .
• Hence wavelength .
• Speed .
• Therefore the speed of sound in the medium is }.