11 physics oscillations and waves test 03 answer 97vd

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CBSE TEST PAPER-03

CLASS - XI PHYSICS (Oscillations and Waves)

Topic: - Waves

Ans 01. This is because no material medium is present over a long distance between earth

and planets and is absence of material medium for propagation, sound waves

cannot travel.

Ans 02. Because propagation of longitudinal waves through a medium, involves changes in

pressure and volume of air, when compressions and rarefactions are formed.

Ans 03. The two prongs of a tuning fork set each other is resonant vibrations and help to

maintain the vibrations for a longer time.

Ans 04. Length of pipe = L = 20cm = 0.2m

Frequency of nth node = n = 430 Hz

Velocity of sound = v = 340m|s

Now, n of closed pipe is :

( )

( )

2 142 1 340

4304 0.2

430 4 0.22 1340

2 1 1.022 1.02 12 2.02

1.01

n

n v

Ln

n

n

n

n

n

=

=

=

=

= +

=

=

Hence, it will be the first normal mode of vibration, In a pipe, open at both ends we,

have

340 340430 , 430=2 2 0.2 2 0.2

430 2 0.2 0.5

340

n

n v n nSoL

n n

= = =

= =

As n has to be an integer, open organ pipe cannot be in resonance with the source.


Ans 05. No, because the emission of light is a random and rapid phenomenon and instead

of beats we get uniform intensity.

Ans 06. Because as speed of sound in water is roughly four times the sound in air, hence

refractive index u = i 1 0.25 r 4

Sin VaSin Vw

= = =

For, refraction r max = 900, imax=140. Since imax rmax hence, sounds gets reflected in

air only and person deep inside the water cannot hear the sound.

Ans 07. Here, depth of well = S = 78.4m Total time after which splash is heard = 4.23s If t1 = time taken by store to hit the water surface is the well t2 = time taken be splash of sound to reach the top of well. then t1 + t2 = 4.23 sec.

Now, for downward journey of stone; u = O, a = 9.8 m|s2, S = 78.4m, t = t1 = ?

As, s = ut + 12

at2

21

21

21

21

1

2

2

2

178.4 0 9.82

78.4 4.978.44.9

1616 4sec

, 1 2 4.234 4.23

4.23 4.000.23

t

t

t

t

t

Now t tt

t

t s

= +

=

=

=

= =

+ =

+ =

=

=

If V = velocity of sound in air,

V = tan ( ) 78.4 340.87 | .( ) 0.23dis ce s

m stime t

= =


Ans 08. Roaring of a lion produces a sound of low pitch and high intensity whereas buzzing of mosquitoes produces a sound of high pitch and low intensity and hence the two sounds can be differentiated.

Ans 09. Let us consider two wares trains of equal amplitude a and with different frequencies

1 2 and in same direction.

Let displacements are y1 and y2 in time t. w1 = Angular vel. of first ware

w1 = 12pi

1 1

1 1

2 2

2 ty = a Sin 2 t

y a Sin w ty a Sin pi

pi

=

=

Acc. to superposition principle, the resultant displacement y at the same time t is:- y = y1 + y2

y = a [ ]1 2 2 t+Sin 2 tSin pi pi Using Sin C + Sin D = 2 Cos Sin

2 2C D C D +

We get,

( ) ( )

( ) ( )1 2 1 2

1 2 1 2

2 22 cos

2 22 Cos Sin

t ty a Sin

y a t t

pi pi

pi pi

+=

= +

Where, A = 2a Cos ( )1 2 .tpi Now, amplitude A is maximum when,

( )( )

( )

1 2

1 2

1 2

ma x + 1=Cos k , k=0,1,2--

t =

Cost k

k

pi pi

pi pi

+ =

=

i.e resultant intensity of sound will be maximum at times,

( ) ( )1 2 1 21 30, ,t

=

( )1 2 Sin y A tpi = +


Time interval between 2 successive Maximaxs = ( ) ( )1 2 1 21 10 (1)

=

Similarly, A will be minimum,

( )( ) ( )

( ) ( )( )( )

1 2

1 2

1 2

1 2

Cos 0

2 1 , 0,1, 2,32

2 12

2 12

t

Cos t Cos k k

t k

kt

pi

pipi

pipi

=

= + =

= +

+=

i.e. resultant intensity of sound will be minimum at times

( ) ( ) ( )1 2 1 2 1 21 3 5

, ,

2 2 2t

=

Hence time interval between 2 successive minim as are

= ( ) ( ) ( )1 2 1 2 1 23 1 1 (2)

2 2 =

Combining 1) & 2) frequency of beats = ( )1 2

. of beatssec

Noonds

= Difference in frequencies of two sources of sound.

Ans 10. The following are the two cases in which there is no Doppler effect in sound (i.e no change in frequency):-

1) When the source of sound as well as the listener moves in the same direction with the same speed.

2) When one of source | listener is at the centre of the circle and the other is moving on the circle with uniform speed.

Ans 11. Let l1, l2 and l3 be the length of the three parts of the wire and f1, f2 and f3 be their respective frequencies.

Since T and m are fixed quantities, and 2 are constant

12

Tfl m

=


l

1

f = constant

fl

or

=

So, f1 l1 = Constant (1) f2 l2 = Constant (2) f3 l3 = Constant (3) Equating equation 1), 2) & 3) f1 l1 = f2 l2 = f3 l3

Now, l2 = 1 12

f lf

12 1

2

1 1(4)2 2

fl l Givenf

= =

Also, 13 13

fl lf=

( )13 13

1 1

3 3fl l givenf

= =

Now, Total length = 110cm

i.e l1 +l2+l3= 110cm

1 1 1

1

10

1

1 1 1102 3

11 1106

110 6 6011

l l l

l

l cm

+ + =

=

= =

i. e.

12

2

2602

ll

l

=

=

Now,

13 3

60

3 3ll l= =

l1 = 60cm

2 30l cm=

3 20l cm=

11 physics oscillations and waves test 03 answer 97vd

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