HC Verma Solutions: Chapter 15 - Wave Motion and Waves on a String | Physics Class 11 - NEET PDF Download

 Page 1


15.1
SOLUTIONS TO CONCEPTS
CHAPTER 15
1. v = 40 cm/sec
As velocity of a wave is constant location of maximum after 5 sec 
= 40 ? 5 = 200 cm along negative x-axis.
2. Given y = 
2
[(x / a) (t / T)]
Ae
? ?
a) [A] = [M
0
L
1
T
0
], [T] = [M
0
L
0
T
1
]
[a] = [M
0
L
1
T
0
]
b) Wave speed, v = ?/T = a/T [Wave length ? = a]
c) If y = f(t – x/v) ? wave is traveling in positive direction
and if y = f( t + x/v) ? wave is traveling in negative direction
So, y = 
2
[(x / a) (t / T)]
Ae
? ?
= 
2
x
(1/ T) t
a / T
Ae
? ?
? ?
? ?
? ?
= 
2
x
(1/ T) t
v
Ae
? ?
? ?
? ?
? ?
i.e. y = f{t + (x / v)}
d) Wave speed, v = a/T
? Max. of pulse at t = T is (a/T) ? T = a (negative x-axis)
Max. of pulse at t = 2T = (a/T) ? 2T = 2a (along negative x-axis)
So, the wave travels in negative x-direction.
3. At t = 1 sec, s
1
= vt = 10 ? 1 = 10 cm
t = 2 sec, s
2
= vt = 10 ? 2 = 20 cm
t = 3 sec, s
3
= vt = 10 ? 3 = 30 cm
4. The pulse is given by, y = [(a
3
) / {(x – vt)
2
+ a
2
}]
a = 5 mm = 0.5 cm, v = 20 cm/s
At t = 0s, y = a
3
/ (x
2
+ a
2
)
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
similarly, at t = 1 s, y = a
3
/ {(x – v)
2
+ a
2
}
and at t = 2 s, y = a
3
/ {(x – 2v)
2
+ a
2
}
5. At x = 0, f(t) = a sin (t/T)
Wave speed = v
? ? = wavelength = vT (T = Time period)
So, general equation of wave
Y = A sin [(t/T) – (x/vT)] [because y = f((t/T) – (x/ ?)) ?
6. At t = 0, g(x) = A sin (x/a)
a) [M
0
L
1
T
0
] = [L]
a = [M
0
L
1
T
0
] = [L] 
b) Wave speed = v
? Time period, T = a/v (a = wave length = ?)
? General equation of wave
y = A sin {(x/a) – t/(a/v)}
= A sin {(x – vt) / a} ?
7. At t = t
0
, g(x, t
0
) = A sin (x/a) …(1)
For a wave traveling in the positive x-direction, the general equation is given by
y = 
x t
f
a T
? ?
?
? ?
? ?
Putting t = –t
0
and comparing with equation (1), we get
? g(x, 0) = A sin {(x/a) + (t
0
/T)}
? g(x, t) = A sin {(x/a) + (t
0
/T) – (t/T)}
x
y
Page 2


15.1
SOLUTIONS TO CONCEPTS
CHAPTER 15
1. v = 40 cm/sec
As velocity of a wave is constant location of maximum after 5 sec 
= 40 ? 5 = 200 cm along negative x-axis.
2. Given y = 
2
[(x / a) (t / T)]
Ae
? ?
a) [A] = [M
0
L
1
T
0
], [T] = [M
0
L
0
T
1
]
[a] = [M
0
L
1
T
0
]
b) Wave speed, v = ?/T = a/T [Wave length ? = a]
c) If y = f(t – x/v) ? wave is traveling in positive direction
and if y = f( t + x/v) ? wave is traveling in negative direction
So, y = 
2
[(x / a) (t / T)]
Ae
? ?
= 
2
x
(1/ T) t
a / T
Ae
? ?
? ?
? ?
? ?
= 
2
x
(1/ T) t
v
Ae
? ?
? ?
? ?
? ?
i.e. y = f{t + (x / v)}
d) Wave speed, v = a/T
? Max. of pulse at t = T is (a/T) ? T = a (negative x-axis)
Max. of pulse at t = 2T = (a/T) ? 2T = 2a (along negative x-axis)
So, the wave travels in negative x-direction.
3. At t = 1 sec, s
1
= vt = 10 ? 1 = 10 cm
t = 2 sec, s
2
= vt = 10 ? 2 = 20 cm
t = 3 sec, s
3
= vt = 10 ? 3 = 30 cm
4. The pulse is given by, y = [(a
3
) / {(x – vt)
2
+ a
2
}]
a = 5 mm = 0.5 cm, v = 20 cm/s
At t = 0s, y = a
3
/ (x
2
+ a
2
)
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
similarly, at t = 1 s, y = a
3
/ {(x – v)
2
+ a
2
}
and at t = 2 s, y = a
3
/ {(x – 2v)
2
+ a
2
}
5. At x = 0, f(t) = a sin (t/T)
Wave speed = v
? ? = wavelength = vT (T = Time period)
So, general equation of wave
Y = A sin [(t/T) – (x/vT)] [because y = f((t/T) – (x/ ?)) ?
6. At t = 0, g(x) = A sin (x/a)
a) [M
0
L
1
T
0
] = [L]
a = [M
0
L
1
T
0
] = [L] 
b) Wave speed = v
? Time period, T = a/v (a = wave length = ?)
? General equation of wave
y = A sin {(x/a) – t/(a/v)}
= A sin {(x – vt) / a} ?
7. At t = t
0
, g(x, t
0
) = A sin (x/a) …(1)
For a wave traveling in the positive x-direction, the general equation is given by
y = 
x t
f
a T
? ?
?
? ?
? ?
Putting t = –t
0
and comparing with equation (1), we get
? g(x, 0) = A sin {(x/a) + (t
0
/T)}
? g(x, t) = A sin {(x/a) + (t
0
/T) – (t/T)}
x
y
Chapter 15
15.2
As T = a/v (a = wave length, v = speed of the wave)
? y = 
0
t x t
A sin
a (a / v) (a / v)
? ?
? ?
? ?
? ?
= 
0
x v(t t)
A sin
a
? ? ? ?
? ?
? ?
? y = 
0
x v(t t )
A sin
a
? ? ? ?
? ?
? ?
8. The equation of the wave is given by
y = (0.1 mm) sin [(31.4 m
–1
)x +(314 s
–1
)t] y = r sin {(2 ?x / ?)} + ?t)
a) Negative x-direction
b) k = 31.4 m
–1
? 2 ?/ ? = 31.4 ? ? = 2 ?/31.4 = 0.2 mt = 20 cm
Again, ? = 314 s
–1
? 2 ?f = 314 ? f = 314 / 2 ? = 314 / (2 ? (3/14)} = 50 sec
–1
? wave speed, v = ?f = 20 ? 50 = 1000 cm/s
c) Max. displacement = 0.10 mm
? Max. velocity = a ? = 0.1 ? 10
–1
? 314 = 3.14 cm/sec. ?
9. Wave speed, v = 20 m/s
A = 0.20 cm
? ? = 2 cm ?
a) Equation of wave along the x-axis
y = A sin (kx – wt)
? k = 2 ?/ ? = 2 ?/2 = ? cm
–1
T = ?/v = 2/2000 = 1/1000 sec = 10
–3
sec
? ? = 2 ?/T = 2 ? ? 10
–3
sec
–1
So, the wave equation is,
? y = (0.2 cm)sin[( ? cm
–1
)x – (2 ? ? 10
3
sec
–1
)t]
b) At x = 2 cm, and t = 0,
? y = (0.2 cm) sin ( ?/2) = 0
? v = r ? cos ?x = 0.2 ? 2000 ? ? cos 2 ? = 400 ?
= 400 ? (3.14) = 1256 cm/s
= 400 ? cm/s = 4 ? m/s ?
10. Y = (1 mm) sin ?
x t
2cm 0.01sec
? ?
?
? ?
? ?
a) T = 2 ? 0.01 = 0.02 sec = 20 ms
? = 2 ? 2 = 4 cm
b) v = dy/dt = d/dt [sin 2 ? (x/4 – t/0.02)] = –cos2 ? {x/4) – (t/0.02)} ? 1/(0.02) ?
? v = –50 cos 2 ? {(x/4) – (t/0.02)}
at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0
c) i) at x = 3 cm, t = 0.01 sec
v = –50 cos 2 ? (3/4 – ½) = 0
ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values)
iii) at x = 7 cm, t = 0.01 sec, v = 0
at x = 1 cm and t = 0.011 sec
v = –50 cos 2 ? {(1/4) – (0.011/0.02)} = –50 cos (3 ?/5) = –9.7 cm/sec
(similarly the other two can be calculated) ?
11. Time period, T = 4 ? 5 ms = 20 ? 10
–3
= 2 ? 10
–2
s
? = 2 ? 2 cm = 4 cm ?
frequency, f = 1/T = 1/(2 ? 10
–2
) = 50 s
–1
= 50 Hz
Wave speed = ?f = 4 ? 50 m/s = 2000 m/s = 2 m/s ?
Page 3


15.1
SOLUTIONS TO CONCEPTS
CHAPTER 15
1. v = 40 cm/sec
As velocity of a wave is constant location of maximum after 5 sec 
= 40 ? 5 = 200 cm along negative x-axis.
2. Given y = 
2
[(x / a) (t / T)]
Ae
? ?
a) [A] = [M
0
L
1
T
0
], [T] = [M
0
L
0
T
1
]
[a] = [M
0
L
1
T
0
]
b) Wave speed, v = ?/T = a/T [Wave length ? = a]
c) If y = f(t – x/v) ? wave is traveling in positive direction
and if y = f( t + x/v) ? wave is traveling in negative direction
So, y = 
2
[(x / a) (t / T)]
Ae
? ?
= 
2
x
(1/ T) t
a / T
Ae
? ?
? ?
? ?
? ?
= 
2
x
(1/ T) t
v
Ae
? ?
? ?
? ?
? ?
i.e. y = f{t + (x / v)}
d) Wave speed, v = a/T
? Max. of pulse at t = T is (a/T) ? T = a (negative x-axis)
Max. of pulse at t = 2T = (a/T) ? 2T = 2a (along negative x-axis)
So, the wave travels in negative x-direction.
3. At t = 1 sec, s
1
= vt = 10 ? 1 = 10 cm
t = 2 sec, s
2
= vt = 10 ? 2 = 20 cm
t = 3 sec, s
3
= vt = 10 ? 3 = 30 cm
4. The pulse is given by, y = [(a
3
) / {(x – vt)
2
+ a
2
}]
a = 5 mm = 0.5 cm, v = 20 cm/s
At t = 0s, y = a
3
/ (x
2
+ a
2
)
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
similarly, at t = 1 s, y = a
3
/ {(x – v)
2
+ a
2
}
and at t = 2 s, y = a
3
/ {(x – 2v)
2
+ a
2
}
5. At x = 0, f(t) = a sin (t/T)
Wave speed = v
? ? = wavelength = vT (T = Time period)
So, general equation of wave
Y = A sin [(t/T) – (x/vT)] [because y = f((t/T) – (x/ ?)) ?
6. At t = 0, g(x) = A sin (x/a)
a) [M
0
L
1
T
0
] = [L]
a = [M
0
L
1
T
0
] = [L] 
b) Wave speed = v
? Time period, T = a/v (a = wave length = ?)
? General equation of wave
y = A sin {(x/a) – t/(a/v)}
= A sin {(x – vt) / a} ?
7. At t = t
0
, g(x, t
0
) = A sin (x/a) …(1)
For a wave traveling in the positive x-direction, the general equation is given by
y = 
x t
f
a T
? ?
?
? ?
? ?
Putting t = –t
0
and comparing with equation (1), we get
? g(x, 0) = A sin {(x/a) + (t
0
/T)}
? g(x, t) = A sin {(x/a) + (t
0
/T) – (t/T)}
x
y
Chapter 15
15.2
As T = a/v (a = wave length, v = speed of the wave)
? y = 
0
t x t
A sin
a (a / v) (a / v)
? ?
? ?
? ?
? ?
= 
0
x v(t t)
A sin
a
? ? ? ?
? ?
? ?
? y = 
0
x v(t t )
A sin
a
? ? ? ?
? ?
? ?
8. The equation of the wave is given by
y = (0.1 mm) sin [(31.4 m
–1
)x +(314 s
–1
)t] y = r sin {(2 ?x / ?)} + ?t)
a) Negative x-direction
b) k = 31.4 m
–1
? 2 ?/ ? = 31.4 ? ? = 2 ?/31.4 = 0.2 mt = 20 cm
Again, ? = 314 s
–1
? 2 ?f = 314 ? f = 314 / 2 ? = 314 / (2 ? (3/14)} = 50 sec
–1
? wave speed, v = ?f = 20 ? 50 = 1000 cm/s
c) Max. displacement = 0.10 mm
? Max. velocity = a ? = 0.1 ? 10
–1
? 314 = 3.14 cm/sec. ?
9. Wave speed, v = 20 m/s
A = 0.20 cm
? ? = 2 cm ?
a) Equation of wave along the x-axis
y = A sin (kx – wt)
? k = 2 ?/ ? = 2 ?/2 = ? cm
–1
T = ?/v = 2/2000 = 1/1000 sec = 10
–3
sec
? ? = 2 ?/T = 2 ? ? 10
–3
sec
–1
So, the wave equation is,
? y = (0.2 cm)sin[( ? cm
–1
)x – (2 ? ? 10
3
sec
–1
)t]
b) At x = 2 cm, and t = 0,
? y = (0.2 cm) sin ( ?/2) = 0
? v = r ? cos ?x = 0.2 ? 2000 ? ? cos 2 ? = 400 ?
= 400 ? (3.14) = 1256 cm/s
= 400 ? cm/s = 4 ? m/s ?
10. Y = (1 mm) sin ?
x t
2cm 0.01sec
? ?
?
? ?
? ?
a) T = 2 ? 0.01 = 0.02 sec = 20 ms
? = 2 ? 2 = 4 cm
b) v = dy/dt = d/dt [sin 2 ? (x/4 – t/0.02)] = –cos2 ? {x/4) – (t/0.02)} ? 1/(0.02) ?
? v = –50 cos 2 ? {(x/4) – (t/0.02)}
at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0
c) i) at x = 3 cm, t = 0.01 sec
v = –50 cos 2 ? (3/4 – ½) = 0
ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values)
iii) at x = 7 cm, t = 0.01 sec, v = 0
at x = 1 cm and t = 0.011 sec
v = –50 cos 2 ? {(1/4) – (0.011/0.02)} = –50 cos (3 ?/5) = –9.7 cm/sec
(similarly the other two can be calculated) ?
11. Time period, T = 4 ? 5 ms = 20 ? 10
–3
= 2 ? 10
–2
s
? = 2 ? 2 cm = 4 cm ?
frequency, f = 1/T = 1/(2 ? 10
–2
) = 50 s
–1
= 50 Hz
Wave speed = ?f = 4 ? 50 m/s = 2000 m/s = 2 m/s ?
Chapter 15
15.3
12. Given that, v = 200 m/s
a) Amplitude, A = 1 mm
b) Wave length, ? = 4 cm
c) wave number, n = 2 ?/ ? = (2 ? 3.14)/4 = 1.57 cm
–1
(wave number = k)
d) frequency, f = 1/T = (26/ ?)/20 = 20/4 = 5 Hz
(where time period T = ?/v) ?
13. Wave speed = v = 10 m/sec
Time period = T = 20 ms = 20 ? 10
–3
= 2 ? 10
–2
sec
a) wave length, ? = vT = 10 ? 2 ? 10
–2
= 0.2 m = 20 cm
b) wave length, ? = 20 cm
? phase diff
n
= (2 ?/ ?) x = (2 ? / 20) ? 10 = ? rad
? y
1
= a sin ( ?t – kx)   ? 1.5 = a sin ( ?t – kx)
So, the displacement of the particle at a distance x = 10 cm.
[ ? = 
2 x 2 10
20
? ? ?
? ? ?
?
] is given by
y
2
= a sin ( ?t – kx + ?) ? –a sin( ?t – kx) = –1.5 mm
? displacement = –1.5 mm ?
14. mass = 5 g, length l = 64 cm
? mass per unit length = m = 5/64 g/cm
? Tension, T = 8N = 8 ? 10
5
dyne
V = 
5
(T /m) (8 10 64)/ 5 3200 ? ? ? ? cm/s = 32 m/s
15.
a) Velocity of the wave, v = 
5
(T /m) (16 10 )/ 0.4 2000 ? ? ? cm/sec
? Time taken to reach to the other end = 20/2000 = 0.01 sec
Time taken to see the pulse again in the original position = 0.01 ? 2 = 0.02 sec
b) At t = 0.01 s, there will be a ‘though’ at the right end as it is reflected.
16. The crest reflects as a crest here, as the wire is traveling from denser to rarer medium. 
? phase change = 0
a) To again original shape distance travelled by the wave S = 20 + 20 = 40 cm.
Wave speed, v = 20 m/s ? time = s/v = 40/20 = 2 sec
b) The wave regains its shape, after traveling a periodic distance = 2 ?30 = 60 cm
? Time period = 60/20 = 3 sec.
c) Frequency, n = (1/3 sec
–1
)
n = (1/2l) (T /m) m = mass per unit length = 0.5 g/cm
? 1/3 = 1/(2 ? 30) (T / 0.5)
? T = 400 ? 0.5 = 200 dyne = 2 ? 10
–3
Newton.
17. Let v
1
= velocity in the 1
st
string
? v
1
= 
1
(T /m )
Because m
1
= mass per unit length = ( ?
1
a
1
l
1
/ l
1
) = ?
1
a
1
where a
1
= Area of cross section
? v
1
= 
1 1
(T / a ) ? …(1)
Let v
2
= velocity in the second string
? v
2
= 
2
(T /m )
? v
2
= 
2 2
(T / a ) ? …(2)
Given that, v
1
= 2v
2
?
1 1
(T / a ) ? = 2
2 2
(T / a ) ? ? (T/a
1
?
1
) = 4(T/a
2
?
2
)
? ?
1
/ ?
2
= 1/4 ? ?
1
: ?
2
= 1 : 4 (because a
1
= a
2
) ?
20 cm
30 cm
Page 4


15.1
SOLUTIONS TO CONCEPTS
CHAPTER 15
1. v = 40 cm/sec
As velocity of a wave is constant location of maximum after 5 sec 
= 40 ? 5 = 200 cm along negative x-axis.
2. Given y = 
2
[(x / a) (t / T)]
Ae
? ?
a) [A] = [M
0
L
1
T
0
], [T] = [M
0
L
0
T
1
]
[a] = [M
0
L
1
T
0
]
b) Wave speed, v = ?/T = a/T [Wave length ? = a]
c) If y = f(t – x/v) ? wave is traveling in positive direction
and if y = f( t + x/v) ? wave is traveling in negative direction
So, y = 
2
[(x / a) (t / T)]
Ae
? ?
= 
2
x
(1/ T) t
a / T
Ae
? ?
? ?
? ?
? ?
= 
2
x
(1/ T) t
v
Ae
? ?
? ?
? ?
? ?
i.e. y = f{t + (x / v)}
d) Wave speed, v = a/T
? Max. of pulse at t = T is (a/T) ? T = a (negative x-axis)
Max. of pulse at t = 2T = (a/T) ? 2T = 2a (along negative x-axis)
So, the wave travels in negative x-direction.
3. At t = 1 sec, s
1
= vt = 10 ? 1 = 10 cm
t = 2 sec, s
2
= vt = 10 ? 2 = 20 cm
t = 3 sec, s
3
= vt = 10 ? 3 = 30 cm
4. The pulse is given by, y = [(a
3
) / {(x – vt)
2
+ a
2
}]
a = 5 mm = 0.5 cm, v = 20 cm/s
At t = 0s, y = a
3
/ (x
2
+ a
2
)
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
similarly, at t = 1 s, y = a
3
/ {(x – v)
2
+ a
2
}
and at t = 2 s, y = a
3
/ {(x – 2v)
2
+ a
2
}
5. At x = 0, f(t) = a sin (t/T)
Wave speed = v
? ? = wavelength = vT (T = Time period)
So, general equation of wave
Y = A sin [(t/T) – (x/vT)] [because y = f((t/T) – (x/ ?)) ?
6. At t = 0, g(x) = A sin (x/a)
a) [M
0
L
1
T
0
] = [L]
a = [M
0
L
1
T
0
] = [L] 
b) Wave speed = v
? Time period, T = a/v (a = wave length = ?)
? General equation of wave
y = A sin {(x/a) – t/(a/v)}
= A sin {(x – vt) / a} ?
7. At t = t
0
, g(x, t
0
) = A sin (x/a) …(1)
For a wave traveling in the positive x-direction, the general equation is given by
y = 
x t
f
a T
? ?
?
? ?
? ?
Putting t = –t
0
and comparing with equation (1), we get
? g(x, 0) = A sin {(x/a) + (t
0
/T)}
? g(x, t) = A sin {(x/a) + (t
0
/T) – (t/T)}
x
y
Chapter 15
15.2
As T = a/v (a = wave length, v = speed of the wave)
? y = 
0
t x t
A sin
a (a / v) (a / v)
? ?
? ?
? ?
? ?
= 
0
x v(t t)
A sin
a
? ? ? ?
? ?
? ?
? y = 
0
x v(t t )
A sin
a
? ? ? ?
? ?
? ?
8. The equation of the wave is given by
y = (0.1 mm) sin [(31.4 m
–1
)x +(314 s
–1
)t] y = r sin {(2 ?x / ?)} + ?t)
a) Negative x-direction
b) k = 31.4 m
–1
? 2 ?/ ? = 31.4 ? ? = 2 ?/31.4 = 0.2 mt = 20 cm
Again, ? = 314 s
–1
? 2 ?f = 314 ? f = 314 / 2 ? = 314 / (2 ? (3/14)} = 50 sec
–1
? wave speed, v = ?f = 20 ? 50 = 1000 cm/s
c) Max. displacement = 0.10 mm
? Max. velocity = a ? = 0.1 ? 10
–1
? 314 = 3.14 cm/sec. ?
9. Wave speed, v = 20 m/s
A = 0.20 cm
? ? = 2 cm ?
a) Equation of wave along the x-axis
y = A sin (kx – wt)
? k = 2 ?/ ? = 2 ?/2 = ? cm
–1
T = ?/v = 2/2000 = 1/1000 sec = 10
–3
sec
? ? = 2 ?/T = 2 ? ? 10
–3
sec
–1
So, the wave equation is,
? y = (0.2 cm)sin[( ? cm
–1
)x – (2 ? ? 10
3
sec
–1
)t]
b) At x = 2 cm, and t = 0,
? y = (0.2 cm) sin ( ?/2) = 0
? v = r ? cos ?x = 0.2 ? 2000 ? ? cos 2 ? = 400 ?
= 400 ? (3.14) = 1256 cm/s
= 400 ? cm/s = 4 ? m/s ?
10. Y = (1 mm) sin ?
x t
2cm 0.01sec
? ?
?
? ?
? ?
a) T = 2 ? 0.01 = 0.02 sec = 20 ms
? = 2 ? 2 = 4 cm
b) v = dy/dt = d/dt [sin 2 ? (x/4 – t/0.02)] = –cos2 ? {x/4) – (t/0.02)} ? 1/(0.02) ?
? v = –50 cos 2 ? {(x/4) – (t/0.02)}
at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0
c) i) at x = 3 cm, t = 0.01 sec
v = –50 cos 2 ? (3/4 – ½) = 0
ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values)
iii) at x = 7 cm, t = 0.01 sec, v = 0
at x = 1 cm and t = 0.011 sec
v = –50 cos 2 ? {(1/4) – (0.011/0.02)} = –50 cos (3 ?/5) = –9.7 cm/sec
(similarly the other two can be calculated) ?
11. Time period, T = 4 ? 5 ms = 20 ? 10
–3
= 2 ? 10
–2
s
? = 2 ? 2 cm = 4 cm ?
frequency, f = 1/T = 1/(2 ? 10
–2
) = 50 s
–1
= 50 Hz
Wave speed = ?f = 4 ? 50 m/s = 2000 m/s = 2 m/s ?
Chapter 15
15.3
12. Given that, v = 200 m/s
a) Amplitude, A = 1 mm
b) Wave length, ? = 4 cm
c) wave number, n = 2 ?/ ? = (2 ? 3.14)/4 = 1.57 cm
–1
(wave number = k)
d) frequency, f = 1/T = (26/ ?)/20 = 20/4 = 5 Hz
(where time period T = ?/v) ?
13. Wave speed = v = 10 m/sec
Time period = T = 20 ms = 20 ? 10
–3
= 2 ? 10
–2
sec
a) wave length, ? = vT = 10 ? 2 ? 10
–2
= 0.2 m = 20 cm
b) wave length, ? = 20 cm
? phase diff
n
= (2 ?/ ?) x = (2 ? / 20) ? 10 = ? rad
? y
1
= a sin ( ?t – kx)   ? 1.5 = a sin ( ?t – kx)
So, the displacement of the particle at a distance x = 10 cm.
[ ? = 
2 x 2 10
20
? ? ?
? ? ?
?
] is given by
y
2
= a sin ( ?t – kx + ?) ? –a sin( ?t – kx) = –1.5 mm
? displacement = –1.5 mm ?
14. mass = 5 g, length l = 64 cm
? mass per unit length = m = 5/64 g/cm
? Tension, T = 8N = 8 ? 10
5
dyne
V = 
5
(T /m) (8 10 64)/ 5 3200 ? ? ? ? cm/s = 32 m/s
15.
a) Velocity of the wave, v = 
5
(T /m) (16 10 )/ 0.4 2000 ? ? ? cm/sec
? Time taken to reach to the other end = 20/2000 = 0.01 sec
Time taken to see the pulse again in the original position = 0.01 ? 2 = 0.02 sec
b) At t = 0.01 s, there will be a ‘though’ at the right end as it is reflected.
16. The crest reflects as a crest here, as the wire is traveling from denser to rarer medium. 
? phase change = 0
a) To again original shape distance travelled by the wave S = 20 + 20 = 40 cm.
Wave speed, v = 20 m/s ? time = s/v = 40/20 = 2 sec
b) The wave regains its shape, after traveling a periodic distance = 2 ?30 = 60 cm
? Time period = 60/20 = 3 sec.
c) Frequency, n = (1/3 sec
–1
)
n = (1/2l) (T /m) m = mass per unit length = 0.5 g/cm
? 1/3 = 1/(2 ? 30) (T / 0.5)
? T = 400 ? 0.5 = 200 dyne = 2 ? 10
–3
Newton.
17. Let v
1
= velocity in the 1
st
string
? v
1
= 
1
(T /m )
Because m
1
= mass per unit length = ( ?
1
a
1
l
1
/ l
1
) = ?
1
a
1
where a
1
= Area of cross section
? v
1
= 
1 1
(T / a ) ? …(1)
Let v
2
= velocity in the second string
? v
2
= 
2
(T /m )
? v
2
= 
2 2
(T / a ) ? …(2)
Given that, v
1
= 2v
2
?
1 1
(T / a ) ? = 2
2 2
(T / a ) ? ? (T/a
1
?
1
) = 4(T/a
2
?
2
)
? ?
1
/ ?
2
= 1/4 ? ?
1
: ?
2
= 1 : 4 (because a
1
= a
2
) ?
20 cm
30 cm
Chapter 15
15.4
18. m = mass per unit length = 1.2 ? 10
–4
kg/mt
Y = (0.02m) sin [(1.0 m
–1
)x + (30 s
–1
)t]
Here, k = 1 m
–1
= 2 ?/ ?
? = 30 s
–1
= 2 ?f
? velocity of the wave in the stretched string
v = ?f = ?/k = 30/I = 30 m/s
? v = T /m ?
4
30 (T /1.2) 10 N)
?
?
? T = 10.8 ? 10
–2
N ? T = 1.08 ? 10
–1
Newton. ?
19. Amplitude, A = 1 cm, Tension T = 90 N
Frequency, f = 200/2 = 100 Hz
Mass per unit length, m = 0.1 kg/mt
a) ? V = T /m = 30 m/s
? = V/f = 30/100 = 0.3 m = 30 cm
b) The wave equation y = (1 cm) cos 2 ? (t/0.01 s) – (x/30 cm)
[because at x = 0, displacement is maximum]
c) y = 1 cos 2 ?(x/30 – t/0.01)
? v = dy/dt = (1/0.01)2 ??sin 2 ? {(x/30) – (t/0.01)}
a = dv/dt = – {4 ?
2
/ (0.01)
2
} cos 2 ? {(x/30) – (t/0.01)}
When, x = 50 cm, t = 10 ms = 10 ? 10
–3
s
x = (2 ? / 0.01) sin 2 ? {(5/3) – (0.01/0.01)}
= (p/0.01) sin (2 ? ? 2 / 3) = (1/0.01) sin (4 ?/3) = –200 ? sin ( ?/3) = –200 ?x ( 3 / 2)
= 544 cm/s = 5.4 m/s
Similarly
a = {4 ?
2
/ (0.01)
2
} cos 2 ? {(5/3) – 1}
= 4 ?
2
? 10
4
? ½ ? 2 ? 10
5
cm/s
2
? 2 km/s
2
20. l = 40 cm, mass = 10 g
? mass per unit length, m = 10 / 40 = 1/4 (g/cm)
spring constant K = 160 N/m
deflection = x = 1 cm = 0.01 m
? T = kx = 160 ? 0.01 = 1.6 N = 16 ? 10
4
dyne
Again v = (T /m) = 
4
(16 10 /(1/ 4) ? = 8 ? 10
2
cm/s = 800 cm/s
? Time taken by the pulse to reach the spring
t = 40/800 = 1/20 = 0/05 sec.
21. m
1
= m
2
= 3.2 kg
mass per unit length of AB = 10 g/mt = 0.01 kg.mt
mass per unit length of CD = 8 g/mt = 0.008 kg/mt
for the string CD, T = 3.2 ? g
? v = (T /m) = 
3
(3.2 10)/ 0.008 (32 10 )/ 8 ? ? ? = 2 10 10 ? = 20 ? 3.14 = 63 m/s
for the string AB, T = 2 ? 3.2 g = 6.4 ? g = 64 N
? v = (T /m) = (64 / 0.01) 6400 ? = 80 m/s
22. Total length of string 2 + 0.25 = 2.25 mt
Mass per unit length m = 
3
4.5 10
2.25
?
?
= 2 ? 10
–3
kg/m
T = 2g = 20 N
Wave speed, v = (T /m) = 
3 4
20 /(2 10 ) 10
?
? ? = 10
2
m/s = 100 m/s
Time taken to reach the pully, t = (s/v) = 2/100 = 0.02 sec.
23. m = 19.2 ? 10
–3
kg/m
from the freebody diagram,
T – 4g – 4a = 0
? T = 4(a + g) = 48 N
wave speed, v = (T /m) = 50 m/s
D
C
B
A
m 2
m 1
2mt
2g
T
25 cm
2kg
4a
a = 2 m/s
2
4 kg
4g
Page 5


15.1
SOLUTIONS TO CONCEPTS
CHAPTER 15
1. v = 40 cm/sec
As velocity of a wave is constant location of maximum after 5 sec 
= 40 ? 5 = 200 cm along negative x-axis.
2. Given y = 
2
[(x / a) (t / T)]
Ae
? ?
a) [A] = [M
0
L
1
T
0
], [T] = [M
0
L
0
T
1
]
[a] = [M
0
L
1
T
0
]
b) Wave speed, v = ?/T = a/T [Wave length ? = a]
c) If y = f(t – x/v) ? wave is traveling in positive direction
and if y = f( t + x/v) ? wave is traveling in negative direction
So, y = 
2
[(x / a) (t / T)]
Ae
? ?
= 
2
x
(1/ T) t
a / T
Ae
? ?
? ?
? ?
? ?
= 
2
x
(1/ T) t
v
Ae
? ?
? ?
? ?
? ?
i.e. y = f{t + (x / v)}
d) Wave speed, v = a/T
? Max. of pulse at t = T is (a/T) ? T = a (negative x-axis)
Max. of pulse at t = 2T = (a/T) ? 2T = 2a (along negative x-axis)
So, the wave travels in negative x-direction.
3. At t = 1 sec, s
1
= vt = 10 ? 1 = 10 cm
t = 2 sec, s
2
= vt = 10 ? 2 = 20 cm
t = 3 sec, s
3
= vt = 10 ? 3 = 30 cm
4. The pulse is given by, y = [(a
3
) / {(x – vt)
2
+ a
2
}]
a = 5 mm = 0.5 cm, v = 20 cm/s
At t = 0s, y = a
3
/ (x
2
+ a
2
)
The graph between y and x can be plotted by taking different values of x.
(left as exercise for the student)
similarly, at t = 1 s, y = a
3
/ {(x – v)
2
+ a
2
}
and at t = 2 s, y = a
3
/ {(x – 2v)
2
+ a
2
}
5. At x = 0, f(t) = a sin (t/T)
Wave speed = v
? ? = wavelength = vT (T = Time period)
So, general equation of wave
Y = A sin [(t/T) – (x/vT)] [because y = f((t/T) – (x/ ?)) ?
6. At t = 0, g(x) = A sin (x/a)
a) [M
0
L
1
T
0
] = [L]
a = [M
0
L
1
T
0
] = [L] 
b) Wave speed = v
? Time period, T = a/v (a = wave length = ?)
? General equation of wave
y = A sin {(x/a) – t/(a/v)}
= A sin {(x – vt) / a} ?
7. At t = t
0
, g(x, t
0
) = A sin (x/a) …(1)
For a wave traveling in the positive x-direction, the general equation is given by
y = 
x t
f
a T
? ?
?
? ?
? ?
Putting t = –t
0
and comparing with equation (1), we get
? g(x, 0) = A sin {(x/a) + (t
0
/T)}
? g(x, t) = A sin {(x/a) + (t
0
/T) – (t/T)}
x
y
Chapter 15
15.2
As T = a/v (a = wave length, v = speed of the wave)
? y = 
0
t x t
A sin
a (a / v) (a / v)
? ?
? ?
? ?
? ?
= 
0
x v(t t)
A sin
a
? ? ? ?
? ?
? ?
? y = 
0
x v(t t )
A sin
a
? ? ? ?
? ?
? ?
8. The equation of the wave is given by
y = (0.1 mm) sin [(31.4 m
–1
)x +(314 s
–1
)t] y = r sin {(2 ?x / ?)} + ?t)
a) Negative x-direction
b) k = 31.4 m
–1
? 2 ?/ ? = 31.4 ? ? = 2 ?/31.4 = 0.2 mt = 20 cm
Again, ? = 314 s
–1
? 2 ?f = 314 ? f = 314 / 2 ? = 314 / (2 ? (3/14)} = 50 sec
–1
? wave speed, v = ?f = 20 ? 50 = 1000 cm/s
c) Max. displacement = 0.10 mm
? Max. velocity = a ? = 0.1 ? 10
–1
? 314 = 3.14 cm/sec. ?
9. Wave speed, v = 20 m/s
A = 0.20 cm
? ? = 2 cm ?
a) Equation of wave along the x-axis
y = A sin (kx – wt)
? k = 2 ?/ ? = 2 ?/2 = ? cm
–1
T = ?/v = 2/2000 = 1/1000 sec = 10
–3
sec
? ? = 2 ?/T = 2 ? ? 10
–3
sec
–1
So, the wave equation is,
? y = (0.2 cm)sin[( ? cm
–1
)x – (2 ? ? 10
3
sec
–1
)t]
b) At x = 2 cm, and t = 0,
? y = (0.2 cm) sin ( ?/2) = 0
? v = r ? cos ?x = 0.2 ? 2000 ? ? cos 2 ? = 400 ?
= 400 ? (3.14) = 1256 cm/s
= 400 ? cm/s = 4 ? m/s ?
10. Y = (1 mm) sin ?
x t
2cm 0.01sec
? ?
?
? ?
? ?
a) T = 2 ? 0.01 = 0.02 sec = 20 ms
? = 2 ? 2 = 4 cm
b) v = dy/dt = d/dt [sin 2 ? (x/4 – t/0.02)] = –cos2 ? {x/4) – (t/0.02)} ? 1/(0.02) ?
? v = –50 cos 2 ? {(x/4) – (t/0.02)}
at x = 1 and t = 0.01 sec, v = –50 cos 2* [(1/4) – (1/2)] = 0
c) i) at x = 3 cm, t = 0.01 sec
v = –50 cos 2 ? (3/4 – ½) = 0
ii) at x = 5 cm, t = 0.01 sec, v = 0 (putting the values)
iii) at x = 7 cm, t = 0.01 sec, v = 0
at x = 1 cm and t = 0.011 sec
v = –50 cos 2 ? {(1/4) – (0.011/0.02)} = –50 cos (3 ?/5) = –9.7 cm/sec
(similarly the other two can be calculated) ?
11. Time period, T = 4 ? 5 ms = 20 ? 10
–3
= 2 ? 10
–2
s
? = 2 ? 2 cm = 4 cm ?
frequency, f = 1/T = 1/(2 ? 10
–2
) = 50 s
–1
= 50 Hz
Wave speed = ?f = 4 ? 50 m/s = 2000 m/s = 2 m/s ?
Chapter 15
15.3
12. Given that, v = 200 m/s
a) Amplitude, A = 1 mm
b) Wave length, ? = 4 cm
c) wave number, n = 2 ?/ ? = (2 ? 3.14)/4 = 1.57 cm
–1
(wave number = k)
d) frequency, f = 1/T = (26/ ?)/20 = 20/4 = 5 Hz
(where time period T = ?/v) ?
13. Wave speed = v = 10 m/sec
Time period = T = 20 ms = 20 ? 10
–3
= 2 ? 10
–2
sec
a) wave length, ? = vT = 10 ? 2 ? 10
–2
= 0.2 m = 20 cm
b) wave length, ? = 20 cm
? phase diff
n
= (2 ?/ ?) x = (2 ? / 20) ? 10 = ? rad
? y
1
= a sin ( ?t – kx)   ? 1.5 = a sin ( ?t – kx)
So, the displacement of the particle at a distance x = 10 cm.
[ ? = 
2 x 2 10
20
? ? ?
? ? ?
?
] is given by
y
2
= a sin ( ?t – kx + ?) ? –a sin( ?t – kx) = –1.5 mm
? displacement = –1.5 mm ?
14. mass = 5 g, length l = 64 cm
? mass per unit length = m = 5/64 g/cm
? Tension, T = 8N = 8 ? 10
5
dyne
V = 
5
(T /m) (8 10 64)/ 5 3200 ? ? ? ? cm/s = 32 m/s
15.
a) Velocity of the wave, v = 
5
(T /m) (16 10 )/ 0.4 2000 ? ? ? cm/sec
? Time taken to reach to the other end = 20/2000 = 0.01 sec
Time taken to see the pulse again in the original position = 0.01 ? 2 = 0.02 sec
b) At t = 0.01 s, there will be a ‘though’ at the right end as it is reflected.
16. The crest reflects as a crest here, as the wire is traveling from denser to rarer medium. 
? phase change = 0
a) To again original shape distance travelled by the wave S = 20 + 20 = 40 cm.
Wave speed, v = 20 m/s ? time = s/v = 40/20 = 2 sec
b) The wave regains its shape, after traveling a periodic distance = 2 ?30 = 60 cm
? Time period = 60/20 = 3 sec.
c) Frequency, n = (1/3 sec
–1
)
n = (1/2l) (T /m) m = mass per unit length = 0.5 g/cm
? 1/3 = 1/(2 ? 30) (T / 0.5)
? T = 400 ? 0.5 = 200 dyne = 2 ? 10
–3
Newton.
17. Let v
1
= velocity in the 1
st
string
? v
1
= 
1
(T /m )
Because m
1
= mass per unit length = ( ?
1
a
1
l
1
/ l
1
) = ?
1
a
1
where a
1
= Area of cross section
? v
1
= 
1 1
(T / a ) ? …(1)
Let v
2
= velocity in the second string
? v
2
= 
2
(T /m )
? v
2
= 
2 2
(T / a ) ? …(2)
Given that, v
1
= 2v
2
?
1 1
(T / a ) ? = 2
2 2
(T / a ) ? ? (T/a
1
?
1
) = 4(T/a
2
?
2
)
? ?
1
/ ?
2
= 1/4 ? ?
1
: ?
2
= 1 : 4 (because a
1
= a
2
) ?
20 cm
30 cm
Chapter 15
15.4
18. m = mass per unit length = 1.2 ? 10
–4
kg/mt
Y = (0.02m) sin [(1.0 m
–1
)x + (30 s
–1
)t]
Here, k = 1 m
–1
= 2 ?/ ?
? = 30 s
–1
= 2 ?f
? velocity of the wave in the stretched string
v = ?f = ?/k = 30/I = 30 m/s
? v = T /m ?
4
30 (T /1.2) 10 N)
?
?
? T = 10.8 ? 10
–2
N ? T = 1.08 ? 10
–1
Newton. ?
19. Amplitude, A = 1 cm, Tension T = 90 N
Frequency, f = 200/2 = 100 Hz
Mass per unit length, m = 0.1 kg/mt
a) ? V = T /m = 30 m/s
? = V/f = 30/100 = 0.3 m = 30 cm
b) The wave equation y = (1 cm) cos 2 ? (t/0.01 s) – (x/30 cm)
[because at x = 0, displacement is maximum]
c) y = 1 cos 2 ?(x/30 – t/0.01)
? v = dy/dt = (1/0.01)2 ??sin 2 ? {(x/30) – (t/0.01)}
a = dv/dt = – {4 ?
2
/ (0.01)
2
} cos 2 ? {(x/30) – (t/0.01)}
When, x = 50 cm, t = 10 ms = 10 ? 10
–3
s
x = (2 ? / 0.01) sin 2 ? {(5/3) – (0.01/0.01)}
= (p/0.01) sin (2 ? ? 2 / 3) = (1/0.01) sin (4 ?/3) = –200 ? sin ( ?/3) = –200 ?x ( 3 / 2)
= 544 cm/s = 5.4 m/s
Similarly
a = {4 ?
2
/ (0.01)
2
} cos 2 ? {(5/3) – 1}
= 4 ?
2
? 10
4
? ½ ? 2 ? 10
5
cm/s
2
? 2 km/s
2
20. l = 40 cm, mass = 10 g
? mass per unit length, m = 10 / 40 = 1/4 (g/cm)
spring constant K = 160 N/m
deflection = x = 1 cm = 0.01 m
? T = kx = 160 ? 0.01 = 1.6 N = 16 ? 10
4
dyne
Again v = (T /m) = 
4
(16 10 /(1/ 4) ? = 8 ? 10
2
cm/s = 800 cm/s
? Time taken by the pulse to reach the spring
t = 40/800 = 1/20 = 0/05 sec.
21. m
1
= m
2
= 3.2 kg
mass per unit length of AB = 10 g/mt = 0.01 kg.mt
mass per unit length of CD = 8 g/mt = 0.008 kg/mt
for the string CD, T = 3.2 ? g
? v = (T /m) = 
3
(3.2 10)/ 0.008 (32 10 )/ 8 ? ? ? = 2 10 10 ? = 20 ? 3.14 = 63 m/s
for the string AB, T = 2 ? 3.2 g = 6.4 ? g = 64 N
? v = (T /m) = (64 / 0.01) 6400 ? = 80 m/s
22. Total length of string 2 + 0.25 = 2.25 mt
Mass per unit length m = 
3
4.5 10
2.25
?
?
= 2 ? 10
–3
kg/m
T = 2g = 20 N
Wave speed, v = (T /m) = 
3 4
20 /(2 10 ) 10
?
? ? = 10
2
m/s = 100 m/s
Time taken to reach the pully, t = (s/v) = 2/100 = 0.02 sec.
23. m = 19.2 ? 10
–3
kg/m
from the freebody diagram,
T – 4g – 4a = 0
? T = 4(a + g) = 48 N
wave speed, v = (T /m) = 50 m/s
D
C
B
A
m 2
m 1
2mt
2g
T
25 cm
2kg
4a
a = 2 m/s
2
4 kg
4g
Chapter 15
15.5
24. Let M = mass of the heavy ball 
(m = mass per unit length)
Wave speed, v
1
= (T /m) = (Mg/m) (because T = Mg)
? 60 = (Mg/m) ? Mg/ m = 60
2
…(1)
From the freebody diagram (2),
v
2
= (T'/m)
? v
2
= 
2 2 1/ 4
1/ 2
[(Ma) (Mg) ]
m
?
(because T’ = 
2 2
(Ma) (Mg) ? )
? 62 = 
2 2 1/ 4
1/ 2
[(Ma) (Mg) ]
m
?
?
2 2
(Ma) (Mg)
m
?
= 62
2
…(2)
Eq(1) + Eq(2) ? (Mg/m) ? [m /
2 2
(Ma) (Mg) ? ] = 3600 / 3844
? g / 
2 2
(a g ) ? = 0.936 ? g
2
/ (a
2
+ g
2
) = 0.876
? (a
2
+ 100) 0.876 = 100
? a
2
? 0.876 = 100 – 87.6 = 12.4
? a
2
= 12.4 / 0.876 = 14.15 ? a = 3.76 m/s
2
? Acce
n
of the car = 3.7 m/s
2
25. m = mass per unit length of the string
R = Radius of the loop
? = angular velocity, V = linear velocity of the string
Consider one half of the string as shown in figure.
The half loop experiences cetrifugal force at every point, away from 
centre, which is balanced by tension 2T.
Consider an element of angular part d ? at angle ?. Consider another 
element symmetric to this centrifugal force experienced by the element 
= (mRd ?) ?
2
R.
(…Length of element = Rd ?, mass = mRd ?)
Resolving into rectangular components net force on the two symmetric elements,
DF = 2mR
2
d ??
2
sin ? [horizontal components cancels each other]
So, total F = 
/ 2
2 2
0
2mR sin d
?
? ? ?
?
= 2mR
2
?
2
[– cos ?] ? 2mR
2
?
2
Again, 2T = 2mR
2
?
2
? T = mR
2
?
2
Velocity of transverse vibration V = T /m = ?R = V
So, the speed of the disturbance will be V. ?
26. a) m ? mass per unit of length of string
consider an element at distance ‘x’ from lower end.
Here wt acting down ward = (mx)g = Tension in the string of upper part
Velocity of transverse vibration = v = T /m = (mgx /m) (gx) ?
b) For small displacement dx, dt = dx / (gx)
Total time T = 
L
0
dx / gx (4L / g) ?
?
c) Suppose after time ‘t’ from start the pulse meet the particle at distance y from lower end.
t = 
y
0
dx / gx (4y / g) ?
?
? Distance travelled by the particle in this time is (L – y)
Mg 
(Rest)
T
a
Mg 
(Motion)
T
Ma
? ?
T
(mRd ?)w
2
R ?
? ?
T
c
d ? ?
y
L-y
A
B
T A
T B
x
4xl