Beats MCQ Quiz - Objective Question with Answer for Beats - Download Free PDF

Calculation: 

Let the frequency of the unknown fork be \( f \) Hz.

The beat frequency is given by the absolute difference between the frequencies of the two forks. Initially, the beat frequency is 4 beats/second.

\(\Rightarrow |f - 288| = 4\)

This gives us two possible values for \( f \):

\(f = 288 + 4 = 292\) Hz or \( f = 288 - 4 = 284\) Hz.

When a little wax is applied, the frequency of the unknown fork decreases. Let's denote the new frequency by \( f' \).

Given that the new beat frequency is 2 beats/second:

\(\Rightarrow |f' - 288| = 2\)

Since the frequency decreases when wax is applied:

\(f' = f - \Delta f\)

Considering \( f = 292 \) Hz (since 292 is one of the possible values from the initial condition):

\(f' = 292 - \Delta f\)

And we know:

\(|292 - \Delta f - 288| = 2\)

\(\Rightarrow |292 - \Delta f - 288| = 2\)

\(\Rightarrow |4 - \Delta f| = 2\)

This gives us two possible values for \( \Delta f \):

\(4 - \Delta f = 2 \Rightarrow \Delta f = 2\)

Therefore, \( f' = 292 - 2 = 290 \) Hz, which does not satisfy the beat frequency of 2 beats/second with 288 Hz.

Considering \( f = 284 \) Hz (since 284 is another possible value):

\(f' = 284 - \Delta f\)

And we know:

\(|284 - \Delta f - 288| = 2\)

\(\Rightarrow |-4 - \Delta f| = 2\)

Since this scenario is not physically possible, we discard \( f = 284 \) Hz.

Therefore, the correct frequency of the unknown fork is:

∴ The correct answer is option 2) 292 Hz

Answer : 2

Solution :

For an open pipe, e = 0.6 d

∴ \(\mathrm{d}=\frac{\mathrm{e}}{0.6}\)

∴ \(2 \mathrm{r}=\frac{\mathrm{e}}{0.6}\)

∴ \(\mathrm{r}=\frac{0.8}{1.2}=\frac{2}{3} \mathrm{~cm}\)

Calculation:

We are given that 31 tuning forks are so arranged that every fork gives 4 beats with the next and the last fork has a frequency that is 1.5 times of the first. We need to find the frequency of the last fork.

Let the frequency of the first fork be \( f \) Hz. Since each fork gives 4 beats with the next, the frequency of the second fork will be \( f + 4 \) Hz, the third fork will be \( f + 8 \) Hz, and so on. Therefore, the frequency of the nth fork can be represented as:

f_n = f + (n-1) \cdot 4

For the 31st fork, \( n = 31 \):

f_{31} = f + 30 \cdot 4 = f + 120

We are also given that the frequency of the last fork is 1.5 times the frequency of the first fork:

f + 120 = 1.5f

Solving for \( f \):

120 = 1.5f - f

120 = 0.5f

f = 240

Therefore, the frequency of the last fork is:

f_{31} = f + 120 = 240 + 120 = 360

Final Answer: The frequency of the last fork is 360 Hz.

Hence frequency in given arrangement are as follows

\(n_{21} = (2 \times 72 - 20 \times 3) = 84 \, \text{Hz}\)

\(n' = \left (\dfrac{v + v_{person}} {v - v_{person}} \right ) \times 272 = \left (\dfrac{345 + 5} {345 - 5} \right ) \times 272 = 280 \, Hz\)

\(\Delta n = Number \, of \, beats = 280 - 272 = 8 \, Hz\)

CONCEPT:

• We consider two sine waves of equal amplitudes and directions. The different frequencies and wavelengths of the waves are considered here while the wave speeds are the same.
• Beat frequency: The difference in frequency of two sound sources sounding together is called beat frequency. 
  • The beat frequency is produced when two waves of different frequencies superimpose when they travel in the same path with the same wave speed.

Let a source have frequency f1 and the other source has frequency f2 then the beat frequency is given by:

Beat frequency (f) = |f1 – f2|

CALCULATION:

Given that,

Frequency of sound wave 1, f₁ = 256 Hz

Frequency of sound wave 2, f₂ = 260 Hz

Thus the frequency of beats is 

f = |f₁ -f₂| = |260-256| = 4 Hz

Hence option 4 is correct among all

CONCEPT:

Beats: 

• When two sound waves of nearly same frequencies, travelling in a medium, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time.
• This phenomenon of regular variation in the intensity of sound with time at a particular position is called beats.
• Beat frequency: The number of beats produced per second is called a beat frequency.

​EXPLANATION:​

• From above it is clear that, the phenomenon of wavering of sound intensity when two waves of nearly the same frequencies and amplitudes travelling in the medium, are superimposed on each other is called beats. Therefore option 4 is correct.

CONCEPT:

• We consider two sine waves of equal amplitudes and directions. The different frequencies and wavelengths of the waves are considered here while the wave speeds are the same.
• Beat frequency: The difference in frequency of two sound sources sounding together is called beat frequency. 
  • The beat frequency is produced when two waves of different frequencies superimpose when they travel in the same path with the same wave speed.

Let a source have frequency f1 and the other source has frequency f2 then the beat frequency is given by:

Beat frequency (f) = |f1 – f2|

EXPLANATION:

• The frequency of the beats heard when two sounds of close frequencies 'f1' and 'f2' superimpose is f₂ - f₁.
• So option 4 is correct.

​CONCEPT:

• Beat: The combination of two waves of slightly different frequencies, perceived as a periodic variation whose rate is the difference of the two frequencies.

Beat frequency (b) = |f1 - f2|

Where f1 and f2 are two frequencies.

CALCULATION:

Given that f_A = 258 Hz and f_B = 262 Hz and the frequency of the unknown fork is f

Let it produces b beats with A and 2b with B. So

f_A - f = b .................(i)

f_B - f = 2b  .................(ii)

so from above equations

f_B - f_A = b

b = 262 - 258 = 4 Hz

from equation (i)

f = f_A - b

f = 258 - 4

unknown frequency f = 254 Hz

So the correct answer is option 3.

CONCEPT:

• Beat: The combination of two waves of slightly different frequencies, perceived as a periodic variation whose rate is the difference of the two frequencies.

Beat frequency (b) = |f1 - f₂|

Where f1 and f2 are two frequencies.

CALCULATION:

Given that b = 6 Hz; f₁ = 324 Hz

Beat frequency (b) = |f1 - f₂|

|324 - f₂| = 6

f2 = 318 Hz or 330 Hz

• So both the frequencies are possible.
• Hence the correct answer is option 3.

The correct answer is option 3) i.e. 6 ​Hz.

CONCEPT:

• Doppler Effect: The phenomenon of change in frequency observed when a source of the sound and a listener are moving relative to each other is called the Doppler effect.
  • When the listener and the source are moving away from one another, the sound heard by the listener will have a frequency lower than the frequency of the sound from the source. 
    • The frequency of the sound heard by the listener, f' = \(\frac{v - v_0}{v + v_s} f_0\)
  • When the listener and source are moving towards one another, the sound heard by the listener will have a frequency higher than the frequency of sound from the source.
    • The frequency of the sound heard by the listener, f' = \(\frac{v + v_0}{v - v_s} f_0\)

Where v is the velocity of sound in air, v₀ is the velocity of the listener, v_s is the velocity of the source of the sound, and f₀ is the frequency of sound emitted by the source.

CALCULATION:

Given that:

Velocity of sound in air, v = 320 m/s

Velocity of train, v_s = 4 m/s

Frequency of whistle made by train, f₀ = 240 Hz

Since the observer is stationary, the velocity of listener v₀ = 0 m/s

Case 1: Train moving away from the observer

The frequency of the sound heard by the listener, f₁' = \(\frac{v + v_0}{v + v_s} f_0 = \frac{ 320 + 0}{320 + 4} \times 240 = \)​ 237 Hz 

Case 2: Train moving towards the observer

The frequency of the sound heard by the listener, f₂' = \(\frac{v + v_0}{v - v_s} f_0 = \frac{320 + 0}{320-4} \times 240 =\)​ 243 Hz

Beats = difference between two frequencies = f2' - f1' = 243 - 237 = 6 Hz

Concept:

• When two sound waves of slightly different frequencies, travelling in a medium along the same direction, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time.
• This phenomenon of regular variation in the intensity of sound with time at a particular position is called beats.
• Beat frequency: The number of beats produced per second is called a beat frequency.

Beat frequency = number of beats/sec

• If n₁ and n₂ are the frequency of two sources, then the difference in frequencies of two sources

m = (n₁ - n₂) or (n₂ – n₁)

Where m = beat frequency

Calculation:

Given Fork A = 384 Hz, 6 beats are produced in 2 seconds

Thus beat frequency

\(b = \frac{6}{2} = 3\;Hz\)

Beat frequency  b = ∣f_A​−f_B​∣

∴ 3 = ∣384−f_B​∣

Thus   f_B​ = 384 – 3 = 381 Hz  or   f_B​ = 384 + 3 = 387 Hz

So, The frequency of B can either be 381 Hz or  387 Hz

CONCEPT:

• We consider two sine waves of equal amplitudes and directions. The different frequencies and wavelengths of the waves are considered here while the wave speeds are the same.
• Beat frequency: The difference in frequency of two sound sources sounding together is called beat frequency. 
  • The beat frequency is produced when two waves of different frequencies superimpose when they travel in the same path with the same wave speed.

Let a source have frequency f1 and the other source has frequency f2 then the beat frequency is given by:

Beat frequency (f) = |f1 – f2|

CALCULATION:

Given that:

Frequency of tuning fork (f₁) = 20 Hz

Frequency of sound of chemical plant machine (f₂) = 25 Hz

Beat frequency (f) = |f1 – f2| = |20 - 25| = |- 5| = 5 Hz

Frequency is always positive.

So option 4 is correct.

CONCEPT:

Beats: 

• When two sound waves of nearly same frequencies, travelling in a medium, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time.
• This phenomenon of regular variation in the intensity of sound with time at a particular position is called beats.
• Beat frequency: The number of beats produced per second is called a beat frequency.

​EXPLANATION:​

• Beats are the interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference between the two frequencies.
• The volume varies as the sounds alternately interfere constructively and destructively. Therefore option 2 is correct.

CONCEPT:

• Interference: When two waves superpose to form a resultant wave.
• Diffraction: It is defined as bending of waves around the corner when a wave encounters an obstacle.
• Refraction: It occurs with light, radio waves, etc. When light passing obliquely from one medium to another medium bends.
• Polarization: It is a property of transverse waves. The direction of oscillations becomes perpendicular to the direction of motion.

EXPLANATION:

• Beats: When two sound waves of slightly different frequencies interfere, a repeating sound is heard known as beats.

• The above diagram shows the interference of two sound waves of slightly different frequencies and resulting beat from zero to high amplitude. In the diagram, C.I. means points of constructive interference and D.I. destructive interference.
• A loud voice is heard or a peak in the beat pattern in the diagram, when constructive interference occurs.
• When destructive interference occurs, no sound is heard as in the diagram no displacement on the beat pattern.
• So, The answer is option 3 i.e. Interference.