**The sound from a very high burst of fireworks takes 5 s to arrive at the observer. The burst occurs 1662 m above the observer and travels vertically through two stratifier layers of air, the top one of thickness**

__Question:__d1 at 0∘C and the bottom one of thickness d2 at 20∘C. Then (assume velocity of sound at 0∘C is 330 m/s)

**Options:**

A. | d1=342m | |

B. | d2=1320m | |

C. | d1=1485m | |

D. | d2=342m |

**Answer: Option D**

: D

Time taken is given by

T=t1+t2=d1v1+d2v2V1=v∘0c=330m/sv2=(330+.06t)=342m/sd=1662m∴T=d1330+(d−d1)342=5sd1(342−330)330×342+d342=5s12d1=5(342×330)−330×1662d1=1320md2=342m

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## More Questions on This Topic :

**The frequency changes by 10% as the source approaches a stationary observer with constant speed Vs. What should be the percentage change in frequency as the source recedes from the observer with the same speed? Given that Vs ≪ v(v is the speed of sound in air).**

__Question 1.__- 14.3%
- 20%
- 16.7%
- 10%

**Answer: Option D**

: D

When the source approaches the observer,

f1=f(vv−vs)=f(1−vsv)−1≈f(1+vsv)or(f1−ff)×100=vsv×100=10.........(1)

In the second case, when the source recedes from the observer

f2=f(vv+vs)=f(1+vsv)−1=f(1−vsv)∴(f2−ff)×100=−vsv×100=−10

[from Eq.(i)]

In the first case, observed frequency increases by 10% while in the second case, observed frequency decreases by 10%

**For a sound wave travelling towards +x direction, sinusoidal longitudinal displacement ξ at a certain time is given as a function of x(Figure). If bulk modulus of air is B=5×105N/m2, the variation of pressure excess will be**

__Question 2.__**Answer: Option D**

: D

ξ=Asin(kx−ωt)Pex=−Bdξdx=−BAkcos(kx−ωt)AmplitudeofPexisBAk=(5×105)(10−4)(2π0.2)=5π×102Pa

**In the figure shown, a source of sound of frequency 510 Hz moves with constant velocity = 20 m/s in the direction shown. The wind is blowing at a constant velocity = 20 m/s towards an observer who is at rest at point B. Corresponding to the sound emitted by the source at initial position A, the frequency detected by the observer is equal to (speed of sound relative to air is 330 m/s)**

__Question 3.__- 510 Hz
- 500 Hz
- 525 Hz
- 550 Hz

**Answer: Option C**

: C

Apparent frequency is given by n′=n(u+vw)(u+vw−vscos60∘) =510(330+20)330+20−20cos60∘ =510×350340=525Hz

**Speed of sound wave is v. If a reflector moves towards a stationary source emitting waves of frequency f with speed u, the wavelength of reflected waves will be**

__Question 4.__- v−uv+uf
- v+uvf
- v+uv−uf
- v−uvf

**Answer: Option C**

: C

Apparent frequency for reflector (which will act here as an observer) would be f1=(v+uv)fWhere f is the actual frequency of source. The reflector will now behave as a source. The apparent frequency will now become

f2=(vv−u)f1

Substituting the value of f1we get

f2=(v+uv−u)f

**A band playing music at frequency f is moving towards a wall at a speed vb. A motorist is following the band with a speed vm. If v is the speed of sound, the expression for the beat frequency heard by the motorist is**

__Question 5.__- v+vmv+vbf
- v+vmv−vbf
- 2vb(v+vm)v2−v2bf
- 2vm(v+vb)v2−v2mf

**Answer: Option C**

: C

The motorist receives two sound waves: direct one and that reflected from the wall.

f′=v+vmv+vbf

For reflected sound waves:

Frequency of sound wave reflected from the wall is

f"=vv−vb×f

Frequency of the reflected waves as received by the moving motorist is

f′"=v+vbv−vb×f′−v+vmv+vbf

Therefore, the beat frequency is

f′"=v+vmv−vb×f−v+vmv+vbf

=2vb(v+vm)v2−v2bf

**A source of sound S is moving with a velocity 50 m/s towards a stationary observer. He measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the sound when it is moving away from the observer after crossing him? The velocity of the sound in the medium is 350 m/s.**

__Question 6.__- 750 Hz
- 857 Hz
- 1143 Hz
- 1333 Hz

**Answer: Option A**

: A

When the source is coming to stationary observer,n′=(vv−vs)n

or n=(1000×300350)Hz

When the source is moving away from the stationary observer,

n"=(vv+vs)n=(350350+50)(1000×300350)=750Hz

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