EBD_7156

S-62 DPP/ CP14

DAILY PRACTICE PHYSICS
PROBLEMS SOLUTIONS DPP/CP14
l1 l2 l3 6. (d) Third overtone has a frequency 7 n, which means
1. (a)
7l
110 cm L= = three full loops + one half loop, which would
4
n1 : n2 : n3 = 3 : 2 : 1 make four nodes and four antinodes.
7. (c) Comparing it with y (x, t) = A cos (wt + p/2) cos kx.
1
nµ If kx = p/2, a node occurs ; \ 10 px = p/2 Þ x = 0.05 m
l
If kx = p, an antinode occurs Þ 10px = p
1 1 1
l1 : l 2 : l 3 = : : = 2:3:6 Þ x = 0.1 m
3 2 1
50p
l1 + l 2 + l 3 = 110 Also speed of wave w / k = = 5m / s and
10p
Þ 2x + 3x + 6x = 110 Þ x = 10 l = 2p / k = 2p / 10p = 0.2 m
\ The two bridges should be set at 2x i.e, 20 cm from
l
one end and 6x i.e, 60 cm from the other end. 8. (a) l1 + x = = 22.7 equation (1)
4
2. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx). 3l
Here u = b l2 + x = = 70.2 equation (2)
4
2p 1
k= = 2pc take, = c 5λ
l l l3 + x = equation (3)
4
1 b
\ Velocity of the wave = ul = b = From equation (1) and (2)
c c
dy l 2 - 3 l 1 70 .2 - 68 .1 2 .1
= a 2pb cos 2p (bt – cx) = aw cos (wt – kx) x= = = = 1.05 cm
dt 2 2 2
Maximum particle velocity = aw= a2pb = 2p ab
l3 + x
b 2 1 From equation (2) and (3) =5
given this is 2 ´ i.e. 2pa = or c = l1 + x
c c pa
3. (c) y = 0.25 sin (10 px – 2pt) l 3 = 5 l 1 + 4x = 5 × 22.7 + 4 × 1.05 =117.7 cm
Comparing this equation with the standard wave H
equation A 0.9 km B I
y = asin (kx – wt) 9. (d)
ENGINE L
We get, k = 10p C L
2p Let after 5 sec engine at point C
Þ = 10p Þ l = 0.2 m
l AB BC
t= +
And w = 2p or, 2pv = 2p Þ v = 1Hz. 330 330
The sign inside the bracket is negative, hence the wave
0.9 ´ 1000 BC
travels in + ve x- direction. 5= +
330 330
2
4. (b) Amplitude of reflected wave = ´ 0 .9 = 0 .6 \ BC = 750 m
3 Distance travelled by engine in 5 sec
It would travel along negative direction of x-axis, and = 900 m – 750 m = 150 m
on reflection at a rigid support, there occurs a phase Therefore velocity of engine
change of p. 150 m
5. (c) Velocity of source = 18 km h–1 = 5 m s–1 = = 30 m/s
5sec
(i) S moves towards listener (vS) 10. (a) For fundamental mode,
(ii) listener moves towards source (vL) 1 T
f=
v + vL 2l m
n' = n = 280 Hz , Beats = n' – n = 8.
v - vS
DPP/ CP14 S-63

Taking logarithm on both sides, we get 15. (d) Figure(a) represents a harmonic wave of frequency
æ Tö 7.0 Hz, figure (b) represents a harmonic wave of
æ 1ö
log f = log ç ÷ + log ç ÷ frequency 5.0 Hz. Therefore beat frequency
è 2l ø è mø vs = 7 – 5 = 2.0 Hz.
æ1 ö 1 æTö
= log çè ÷ø + log ç ÷ 16. (a)uµ T
2l 2 èmø
17. (b) In fundamental mode,
æ1ö 1 l
or log f = log ç ÷ + [log T - log m] = l Þ l = 2l
è 2l ø 2
2
Differentiating both sides, we get
v v
df 1 dT \ f = = ....... (1)
= (as l and m are constants) l 2l
f 2 T
dT df A A
Þ = 2´
T f
l 2 l 4
Here df = 6
f = 600 Hz N
l N l/2
dT 2 ´ 6
\ = = 0.02
T 600
11. (b) Frequency received by listener from the rear source,
v -u v -u v v-u A
n¢ = ´n = ´ =
v v l l Fundamental mode Half length dipped
Frequency received by listener from the front source, in water

v+u v v+u In half length dipped in water mode,

n¢¢ = ´ = l l
v l l = Þ l = 2l
No. of beats = n'' – n' 2 4
v+u v -u v + u - v + u 2u v v
- = \ f '= = = f
=
l l
=
l l l 2l
18. (a) Given wave equation is y(x,t)
é vù
é v + v0 ù êv + 5 ú é6ù =e
(- ax +bt +2
2 2
ab xt )
12. (c) n ' = n ê v ú = n ê ú = nê ú
ë û ê v ú ë5û -[( ax )2 + ( b t ) 2 + 2 a x . b t ]
êë úû =e
n' 6 n' - n 6 - 5 -( ax + bt ) 2
= ; = ´ 100 = 20% =e
n 5 n 5 æ ö
2
13. (c) In a closed organ pipe the fundamental frequency is - çè x +
b ÷
tø
e a
u =
u= It is a function of type y = f (x + vt)
4L
320 ms -1 Þ Speed of wave =
b
u= = 80 Hz a
4 ´1 m
In a closed organ pipe only odd harmonics are present. 19. (c) Particle velocity
So, it can resonate with 80 Hz, 240 Hz, 400 Hz, 560 Hz. d é æ x öù
14. (b) v = dt ê x 0 sin 2pç nt - l ÷ú
ë è øû
f f' f" f'
æ xö
S 30 ms
–1
O O 30 ms
–1
S = 2 pnx 0 cos 2pç nt - ÷
è l ø
f ¢ is the apparent frequency received by an observer at
the hill. f ¢¢ is the frequency of the reflected sound as \ Maximum particle velocity = 2pnx 0
heard by driver.
l
v Wave velocity = = nl
f '= f, T
v - 30
2pnx 0 px 0
v + 30 v + 30 360 Given, 2pnx 0 = 4nl Þ l = =
f"= f'= f= ´ 600 4n 2
v v - 30 300 20. (d) As number of beats/sec = diff. in frequencies has to be
= 720 Hz
less than 10, therefore 0 < (n1 – n2) < 10
 EBD_7156
S-64 DPP/ CP14
21. (c) Length of pipe = 85 cm = 0.85m I I
Frequency of oscillations of air column in closed organ or, 1 = 102 or, I2 = 1 .
I2 100
pipe is given by,
Þ Intensity decreases by a factor 100.
(2n - 1)u y(x, t) = 0.005 cos (ax - bt) (Given)
f = 28. (a)
4L Comparing it with the standard equation of wave
(2 n - 1)u y(x, t) = a cos (kx - wt) we get
f = £ 1250
4L
k = a and w = b
(2n - 1) ´ 340
Þ £ 1250 2p 2p
0.85 ´ 4 But k = and w =
Þ 2n – 1 < 12.5 » 6 l T
22. (b) As the source is not moving towards or away from the 2p 2p
Þ = a and =b
observer in a straight line, so the Doppler’s effect will not l T
be observed by the observer. Given that l = 0.08 m and T = 2.0s
23. (c) Frequency of first source with 5 beats/ sec = 100 Hz and 2p 2p
frequency of second source with 5 beats/sec = 205 Hz. \ a= = 25p and b = =p
0.08 2
The frequency of the first source = 100 ± 5 = 105 or 95 Hz.
29. (b) Equation is of stationary wave. Comparing with the
Therefore, frequency of second harmonic of source = 210
standard equation
Hz or 190 Hz. As the second harmonic gives 5 beats/
second with the sound of frequency 205 Hz, therefore, æ 2p ö æ 2p ö
y = 2A sin çè ÷ø t cos çè ÷ø x
frequency of second harmonic source should be 210 Hz T l
or frequency of source = 105 Hz. 2p 2p
= 4.5 or l = = 1.4m
24. (c) Pressure change will be minimum at both ends. In fact, l 4.5
pressure variation is maximum at l/2 because the dis- 30. (b) Waves are kind of disturbances which moves from one
placement node is pressure antinode. place to another without the actual physical transfer of
25. (a) nLast = nFirst + (N – 1)x matter of the medium as a whole. The particles of the
medium only oscillate but do not travel from one place
2n = n + (41 – 1 ) × 5
to another.
Þ nFirst = 200 Hz and nLast = 400 Hz Waves transport energy and the pattern of disturbance
26. (b) Here, T = 0.05 sec, v = 300 ms–1. has information that propagate from one point to
another. Here, wave pattern propagates.
v
Now l = = vT = (300 ´ 0.05)m All our communication essentially depend on
v
transmission of signals through the waves.
or, l = 15 m
v 3v l1 1
Phase of the point at 10 m from the source 31. (b) = , \ =
4 l1 2 l 2 l2 6
2p 2p 4p
= ´x= ´ 10 = rad 32. (c) w1 = 600p, w2 = 604p,
l 15 3
Phase of the point at 15 m from the source f1 = 300 Hz, f2 = 302 Hz
Beat frequency, f2 – f1 = 2 Hz
2p 2p
´x = ´ 15 = 2p rad Þ number of beats in three seconds = 6
l 15
\ The phase difference between the points 33. (b) Given f A = 1800Hz
4 p 2p vt = v
= 2p - = rad fB = 2150 Hz
3 3
æ I2 ö Reflected wave frequency received by A, f A¢ = ?
æI ö
27. (a) We have, L1 = 10log ç 1 ÷ ; L2 = 10 log ç ÷ Applying doppler’s effect of sound,
èI ø 0
è I0 ø
vs f
æI ö æI ö f¢ =
\ L1 – L2 = 10 log ç 1 ÷ - 10log ç 2 ÷ vs - v t
è 0ø
I è I0 ø
æ f ö
æI I ö æI ö here, v t = vs ç1 - A ÷
or, DL = 10 log ç 1 ´ 0 ÷ or, DL = 10log ç 1 ÷ è fB ø
è 0 2ø
I I è I2 ø
æ 1800 ö
= 343 ç 1 - ÷
æI ö æI ö è 2150 ø
or, 20 = 10 log 1 or, 2 = log 1
çè I ÷ø çè I ÷ø vt = 55.8372 m/s
2 2
DPP/ CP14 S-65

Now, for the reflected wave,

1
or, n µ or nl = constant, K
æ v + vt ö l
\ f A¢ = ç s ÷ fA
è vs - v t ø \ n1l1 = K,
n2l2 = K, n3l3 = K
æ 343 + 55.83 ö
=ç ÷ ´ 1800 Also, l = l1 + l2 +l3
è 343 - 55.83 ø
K K K K
= 2499.44 » 2500Hz or, n = n + n + n
34. (a) Standing waves are produced when two waves 1 2 3
propagate in opposite direction 1 1 1 1
or, n = n + n + n
As z 1 & z 2 are propagating in +ve x-axis & 1 2 3
–ve x-axis 2
38. (a) Time taken for two syllables t = sec.
so, z1 + z2 will represent a standing wave. 5
35. (d) Load supported by sonometer wire = 4 kg
2
Tension in sonometer wire = 4 g x + x = v ´ t = 330 ´ \ x = 66 m
5
If m = mass per unit length
g RT
1 T 39. (a) Velocity of sound =
then frequency u = M
2l m
When water vapour are represent in air average
1 4g molecular weight of air decreases and hence velocity
Þ 416 = increases.
2l m
When length is doubled, i.e., l¢ = 2l ᔼ max a + b㡰
a+b
40. (b) = = 49 \ =7
Let new load = L ᔼ min a 㜠 b㡰
a-b
As, u¢ = u a 8 4
= =
7a – 7b = a + b or 6a = 8b or
1 Lg 1 4g b 6 3
\ =
2l ¢ m 2l m 41. (a) By the concept of accoustic, the observer and source
are moving towards each other, each with a velocity of
1 Lg 1 4g 18 m s–1.
Þ =
4l m 2l m 330 + 18
\n ' = ´1000 » 1115 Hz
330 - 18
Þ L = 2 ´ 2 Þ L = 16 kg
42. (c) Compare the given equation with standard form
36. (b) Let the string vibrates in p loops, wavelength of the
pth mode of vibration is given by é 2px 2pt ù
y = r sin ê -
2l ë l T úû
lp =
p 2p 2p 2p
= 3, l = and = 15
æ 4px ö l 3 T
Given, y = 2 sin ç ÷ cos(96 pt)
è 15 ø 2p
T=
é æ 4px ö æ 4px ö 15
or y = 2 êsin ç + 96pt ÷ + sin ç – 96pt ÷
ë è 15 ø è 15 ø l 2p / 3
Comparing it with standard equation, we get Speed of propagation, v = = =5
T 2 p / 15
96 p 4p
u= = 48 Hz and k = 43. (c) The contrast will be maximum, when I1 = I2 i.e.
2p 15
a = b. In that event, Imin = (a – b)2 = 0, where a and b are
1 2 ´ 60 4p the amplitudes of interfering waves.
= ´
48 p 15 ´ 96 p 44. (b) Fundamental frequency of closed organ pipe
Þ p = 16. V
Vc =
l1 l2 l3 4lc
37. (b)
Fundamental frequency of open organ pipe
1 T V
n= V0 =
2l0
2l m
 EBD_7156
S-66 DPP/ CP14
3V w 2 p l 180
Second overtone frequency of open organ pipe = v= = ´ = = 30 m / s
2l0 k T 2p 6
From question, Differentiating (1) w.r.t. t,
V 3V dy
= v= = -60 ´ 180 sin(180 t - 6 x )
4lc 2l0 dt
Þ l0 = 6lc = 6 × 20 = 120 cm vmax = 60 × 180 mm/s
45. (b) y = 60 cos (180t – 6x) ....(1) = 10800 mm/s = 0.0108 m/s
2p v max 0.0108
w = 180, k = 6 Þ =6 = = 3.6 ´ 10 - 4
l v 30