Test: Waves - 3 - Grade 12 MCQ

Because in wave propagation particles only moves in perodic motion, particles get collide with their neighbouring particles then transfer energy to them and comes back to their normal stage. Now next particle which gain energy get into excited state and starts moving periodically and get collide to next adjacent particle and so on. Thus in wave propagation particles only transfer their kinetic energy and momentum. Hence particles does not move therefore there is no flow of matter but there is movement of disturbance. 

For a sinusoidal wave, V = v λ.
V = speed,
v = frequency,
λ = wavelength,
frequency (v) = reciprocal of the time period i.e. v =1/T

Velocity of sound in air= 300m/s =300×100=30000 cm/s
And frequency = 1000 hz
So, wavelength = Velocity/frequency
= 30000/1000= 30
Distance = wavelength/2
=30/2 = 15

As we know, wavelength = speed/frequency
Wavelength 1 = 1m
Wavelength 2 = 1.01m
=> 1 = v/f1
f₁ =  v
And 1.01 = v/f2
f₂ = v/1.01
Now beat is 19 beats/3 secs
f₁ - f₂= 10/3
Solving equations we get
v = 336.6m/s
Hence B is the correct answer.

In open end organ pipe both odd and even harmonics are produced. This is the definition of open end organ pipe.

Explanation:Because mobile communication is a space communication and in space communication basically electromagnetic waves are used (as carrier waves as in case of radio communication) because of the modulation ( frequency, amplitude) operations which can be performed on EM waves. Thus when our friend receives the call, he also receives EM waves which is the carrier of our audio signals.

Resultant displacement of the wave by these three wave is
y=asin2π400t+asin2π401t+asin2π402t
y=a(1+2cos2πt)sin2π401t
So the resultant magnitude a(1+2cos2πt) has a maximum when,
cos2πt=1
or, t=0,1,2...
The time interval between two successive maximum is 1 sec.
So beat frequency is 1sec.

Because longitudinal waves are the mechanical waves that need a medium to propagate such as air, gas, solid etc. but these are not available in vacuum, so this wave can't propagate in vacuum.

Explanation:Electromagnetic waves are transverse waves, they move perpendicular to the direction of propagation of wave ( the direction in which energy is transferred) and EM waves( Electromagnetic waves) can travel in vacuum, thus doesn't require any medium also.

Let the waves be
y1 = Asin(wt)
y2 = Asin(wt + φ)
If φ = 0 or 2nπ then the waves are exactly in phase and the interference is constructive.
Hence A is the correct answer.

Beats = |f1 – f2| where f1 and f2 are the frequencies
Now, f1 = v/λ1 and f2 = v/λ2
So, 4 = |v (1/1- 1/0.01)|
v = 0.0404 m/s

Explanation:Matter waves are also termed as De Broglie waves because they were initially introduced by him. All matters behave like a wave (during motion) and in there propagation there is action of forces due to collision between the particles, some time torque also exists, hence all the matters waves were consider to be a helpful part for understanding quantum.

After waxing the frequency of the tuning fork decreases and so the initial frequency must be higher so as to decrease it below 256 Hz in order to give 4 beats/sec.

Explanation:As in air wave propagates in the form of compression (increase in density of air) and rarefaction (decrease in density of air).

Because, frequency of a wave depend on source
V1/λ1 =V2/λ2 as frequency is constant
V1 - velocity of sound wave in air
V2 - velocity of sound wave in water

Turning fork = 56turning
Beats number=4beats/s
A frequency of the last fork
Number of beats in each second= difference between frequency=4Hz
The frequency of the tuning fork of 56Hz beats3×n
(When n be the frequency of the first beat of the tuning fork)
Now, There are 55 difference between a set of 56 tunning fork.
The difference between two frequencies4Hz
Now, the difference between 56Hz4 and 1st tunning frequencies=3n−n
= 2n
As per equation,
3n−n=2n
2n=55×4
n = 55×4/2
= 110Hz

Explanation:As progressive wave means a wave propagating in some onward direction in the medium

Let us say that we speak syllables at a rate of 2 to 9 per second. So let us say that a syllable takes a minimum of 0.1 sec for a fast speaker. Let us say that a sound pulse (syllable) is emitted starting at t = 0.

The effect of a syllable lasts on the ear for 0.1 sec. So if any echo reaches the year before t = 0.2 sec., then it is mixed with the direct sound present in the ear and so echo is not properly heard.

In this problem, two syllables are repeated in the echo. That is it took about 2 * 0.2 sec ie., 0.4 seconds for the sound to travel to the reflecting surface and come back to the ear.

The distance of the reflecting surface from the person

= 330 m/s * 0.4 sec / 2

= 66 meters.

Explanation:

As capillary & gravity waves are elastic waves or mechanical waves which require medium for their propagation.

Hence they are using the elastic behaviour of water.

Hence

The waves on the surface of water are of two kinds:

capillary waves and gravity waves

c = λν
Therefore, λ = 25 m

Polarisation is a phenomenon of converting unpolarized light to polarized. It is not done by sound waves.

Explanation:

As speed of transverse & longitudinal waves depend on different modulus of elasticity (Young's modulus, Bulk modulus, Modulus of Rigidity) of the medium.

So in the same medium transverse and longitudinal waves

travel with different speeds

The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum (towards the orange/red/infrared/microwave/radio end of the spectrum).