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Waves NEET Questions — Free MCQ Practice

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Two waves having sinusoidal waveforms have different wavelengths and different amplitudes. They will be having :

A.

Same pitch and different intensity

B.

Same quality and different intensity

C.

Different quality and different intensity

D.

Same quality and different pitch

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Explanation

The pitch depends upon the frequency of the source. As the two waves have different amplitude therefore they having different intensity. While quality depends on number of harmonics/overtone produced and their relative intensity

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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 $y_{1} = A \sin (π x / L) \sin ω t$ and energy is E₁, and in another experiment its displacement is $y_{2} = A \sin (2 π x / L) \sin 2 ω t$ and energy is E₂. Then :

A.

E₂ = E₁

B.

E₂ = 2E₁

C.

E₂ = 4E₁

D.

E₂ = 16E₁

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Explanation

Energy (E) ∝ (Amplitude)² (Frequency)²

Amplitude is same in both the cases, but frequency 2ω in the second case is two times the frequency (ω) in the first case. Hence E₂ = 4E₁.

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In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. when this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction :

A.

0.012m

B.

0.025m

C.

0.05m

D.

0.024m

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Explanation

Let x be the end correction then according to question.

$\frac{v}{4 (l_{1} + x)} = \frac{3 v}{4 (l_{2} + x)} \Rightarrow x = 2 .5 c m = 0.023 m$ .

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Two identical stringed instruments have a frequency 100 Hz. If the tension in one of them is increased by 4% and they are sounded together then the number of beats in one second is :

A.

1

B.

8

C.

4

D.

2

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Explanation

Frequency of vibration in tight string

$n = \frac{p}{2 l} \sqrt{\frac{T}{m}} \Rightarrow n \propto \sqrt{T}$ ⇒ $\frac{Δ n}{n} = \frac{Δ T}{2 T} = \frac{1}{2} \times (4 %) = 2 %$

⇒ Number of beats = $Δ n = \frac{2}{100} \times n = \frac{2}{100} \times 100 = 2$

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The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. If the velocity of sound in air is 300 m/sec, the velocity of the source is : (It is given that velocity of source << velocity of sound)

A.

6 m/sec

B.

3 m/sec

C.

1.5 m/sec

D.

12 m/sec

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Explanation

When the source approaches the observer

Apparent frequency $n' = \frac{v}{v - v_{s}} . n = n [\frac{1}{1 - \frac{v_{s}}{v}}]$

= $n {[1 - \frac{v_{s}}{v}]}^{- 1} = n [1 + \frac{v_{s}}{v}]$

(Neglecting higher powers because of v_S << v)

When the source recedes the observed apparent frequency $n^{''} = n [1 - \frac{v_{s}}{v}]$

Given $n' - n'' = \frac{2}{100} n, v = 300 m / s e c$

∴ $\frac{2}{100} n = n [1 + \frac{v_{s}}{v}] - n [1 - \frac{v_{s}}{v}] = n [2 \frac{v_{s}}{v}]$

$\Rightarrow \frac{2}{100} = 2 \frac{v_{s}}{v} \Rightarrow v_{s} = \frac{v}{100} = \frac{300}{100} = 3 m / s e c$

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Two whistles A and B produce notes of frequencies 660 Hz and 596 Hz respectively. There is a listener at the mid-point of the line joining them. Now the whistle B and the listener start moving with speed 30 m/s away from the whistle A. If the speed of sound be 330 m/s, how many beats will be heard by the listener :

A.

2

B.

4

C.

6

D.

8

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Explanation

For observer note of B will not change due to zero relative motion.

Observed frequency of sound produced by A

= $660 \frac{(330 - 30)}{330} = 600 H z$

∴ No. of beats = 600 – 596 = 4

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A source producing the sound of frequency 170 Hz is approaching a stationary observer with a velocity of 17 ms^–1. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air = 340 ms^–1)

A.

0.1m

B.

0.2m

C.

0.4m

D.

0.5m

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Explanation

$λ = \frac{v}{n} = \frac{340}{170} = 2 m, n' = \frac{340}{340 - 17} \times 170 \Rightarrow n' = 178 .9 H z$

Now $λ^{'} = \frac{v}{n^{'}} = \frac{340}{178 .9} = 1 .9$

⇒ $λ - λ^{'} = 2 - 1 .9 = 0 .1$

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An observer moves towards a stationary source of sound with a speed 1/5^th of the speed of sound. The wavelength and frequency of the sound emitted are λ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively :

A.

$1.2 f, λ$

B.

$f, 1 .2 λ$

C.

$0.8 f, 0 .8 λ$

D.

$1 .2 f, 1 .2 λ$

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Explanation

$n' = \frac{v + v_{0}}{v} . f = \frac{v + \frac{v}{5}}{v} . f = 1 .2 f$

and since the source is stationary, so wavelength remains unchanged for observer.

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The equation of displacement of two waves are given as $y_{1} = 10 \sin (3 π t + \frac{π}{3})$ ; $y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$ . Then what is the ratio of their amplitudes ?

A.

1 : 2

B.

2 : 1

C.

1 : 1

D.

None of these

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Explanation

$y_{1} = 10 \sin (3 π t + \frac{π}{3})$ ...(i)

and $y_{2} = 5 [\sin 3 π t + \sqrt{3} \cos 3 π t]$

$= 5 \times 2 [\frac{1}{2} \times \sin 3 π t + \frac{\sqrt{3}}{2} \times \cos 3 π t]$

$= 10 [\cos \frac{π}{3} \sin 3 π t + \sin \frac{π}{3} \cos 3 π t]$

$= 10 [\sin (3 π t + \frac{π}{3})]$ ... (ii)

(âˆµ sin(A + B) = sinA cosB + cosA sinB)

Comparing equation (i) and (ii) we get ratio of amplitude 1 : 1.

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Consider ten identical sources of sound all giving the same frequency but having phase angles which are random. If the average intensity of each source is I₀, the average of resultant intensity I due to all these ten sources will be :

A.

I = 100 I₀

B.

I = 10 I₀

C.

I = I₀

D.

$I = \sqrt{10} I_{0}$

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Explanation

In case of interference of two waves resultant intensity

$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

If Ï• varies randomly with time, so ${(\cos ϕ)}_{a v} = 0$

⇒ $I = I_{1} + I_{2}$

For n identical waves, $I = I_{0} + I_{0} + . . . . . . . = n I_{0}$

Here $I = 10 I_{0} .$

Waves MCQs for NEET — Physics Questions with Answers

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