Chapter 12 : Simple Harmonics Motion - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

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JEE : Chapter 12 : Simple Harmonics Motion - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


12.1
SOLUTIONS TO CONCEPTS 
CHAPTER 12
1. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w = 
T
2 ?
= 
6
2 ?
= 
3
?
sec
–1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ?) = 10 sin ? [y = r sin wt]
Sin ? = 1/2 ? ? = 
6
?
? Equation of displacement x = (10cm) sin ?
?
?
?
?
? ?
3
(ii) At t = 4 second
x = 10 sin 
?
?
?
?
?
? ?
? ?
?
6
4
3
= 10 sin 
?
?
?
?
?
? ? ? ?
6
8
= 10 sin ?
?
?
?
?
? ?
2
3
= 10 sin ?
?
?
?
?
? ?
? ?
2
= - 10 sin ?
?
?
?
?
? ?
2
= -10
Acceleration a = – w
2
x = –
?
?
?
?
?
?
?
?
?
9
2
× (–10) = 10.9 ? 0.11 cm/sec. ?
2. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec
–2
We know that a = ?
2
x 
? ? = 
x
a
= 
02 . 0
10
= 500 = 10 5
T = 
?
? 2
= 
5 10
2 ?
= 
236 . 2 10
14 . 3 2
?
?
= 0.28 seconds.
Again, amplitude r is given by v = ?
?
?
?
?
?
?
?
2 2
x r ?
? v
2
= ?
2
(r
2
– x
2
)
1 = 500 (r
2
– 0.0004)
? r = 0.0489 ? 0.049 m
? r = 4.9 cm. ?
3. r = 10cm
Because, K.E. = P.E.
So (1/2) m ?
2
(r
2
– y
2
) = (1/2) m ?
2
y
2
r
2
– y
2
= y
2
? 2y
2 
= r
2
? y = 
2
r
= 
2
10
= 5 2 cm form the mean position. ?
4. v
max
= 10 cm/sec.
? r ? = 10
? ?
2
= 
2
r
100
…(1)
A
max
= ?
2
r = 50 cm/sec
? ?
2
= 
y
50
= 
r
50
…(2)
Page 2


12.1
SOLUTIONS TO CONCEPTS 
CHAPTER 12
1. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w = 
T
2 ?
= 
6
2 ?
= 
3
?
sec
–1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ?) = 10 sin ? [y = r sin wt]
Sin ? = 1/2 ? ? = 
6
?
? Equation of displacement x = (10cm) sin ?
?
?
?
?
? ?
3
(ii) At t = 4 second
x = 10 sin 
?
?
?
?
?
? ?
? ?
?
6
4
3
= 10 sin 
?
?
?
?
?
? ? ? ?
6
8
= 10 sin ?
?
?
?
?
? ?
2
3
= 10 sin ?
?
?
?
?
? ?
? ?
2
= - 10 sin ?
?
?
?
?
? ?
2
= -10
Acceleration a = – w
2
x = –
?
?
?
?
?
?
?
?
?
9
2
× (–10) = 10.9 ? 0.11 cm/sec. ?
2. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec
–2
We know that a = ?
2
x 
? ? = 
x
a
= 
02 . 0
10
= 500 = 10 5
T = 
?
? 2
= 
5 10
2 ?
= 
236 . 2 10
14 . 3 2
?
?
= 0.28 seconds.
Again, amplitude r is given by v = ?
?
?
?
?
?
?
?
2 2
x r ?
? v
2
= ?
2
(r
2
– x
2
)
1 = 500 (r
2
– 0.0004)
? r = 0.0489 ? 0.049 m
? r = 4.9 cm. ?
3. r = 10cm
Because, K.E. = P.E.
So (1/2) m ?
2
(r
2
– y
2
) = (1/2) m ?
2
y
2
r
2
– y
2
= y
2
? 2y
2 
= r
2
? y = 
2
r
= 
2
10
= 5 2 cm form the mean position. ?
4. v
max
= 10 cm/sec.
? r ? = 10
? ?
2
= 
2
r
100
…(1)
A
max
= ?
2
r = 50 cm/sec
? ?
2
= 
y
50
= 
r
50
…(2)
Chapter 12
12.2
?
2
r
100
= 
r
50
? r = 2 cm.
? ? = 
2
r
100
= 5 sec
2
Again, to find out the positions where the speed is 8m/sec,
v
2
= ?
2
(r
2
– y
2
)
? 64 = 25 ( 4 – y
2
)
? 4 – y
2
= 
25
64
? y
2
= 1.44 ? y = 44 . 1 ? y = ?1.2 cm from mean position.
5. x = (2.0cm)sin [(100s
–1
) t + ( ?/6)]
m = 10g.
a) Amplitude = 2cm.
? = 100 sec
–1
? T = 
100
2 ?
= 
50
?
sec = 0.063 sec.
We know that T = 2 ?
k
m
? T
2
= 4 ?
2
×
k
m
? k = m
T
4
2
2
?
= 10
5
dyne/cm = 100 N/m. [because ? = 
T
2 ?
= 100 sec
–1
]
b) At t = 0
x = 2cm sin ?
?
?
?
?
? ?
6
= 2 × (1/2) = 1 cm. from the mean position.
We know that x = A sin ( ?t + ?)
v = A cos ( ?t + ?)
= 2 × 100 cos (0 + ?/6) = 200 × 
2
3
= 100 3 sec
–1
= 1.73m/s
c) a = – ?
2
x = 100
2
× 1 = 100 m/s
2
?
6. x = 5 sin (20t + ?/3)
a) Max. displacement from the mean position = Amplitude of the particle.
At the extreme position, the velocity becomes ‘0’.
? x = 5 = Amplitude.
? 5 = 5 sin (20t + ?/3)
sin (20t + ?/3) = 1 = sin ( ?/2)
? 20t + ?/3 = ?/2
? t = ?/120 sec., So at ?/120 sec it first comes to rest.
b) a = ?
2
x = ?
2
[5 sin (20t + ?/3)]
For a = 0, 5 sin (20t + ?/3) = 0 ? sin (20t + ?/3) = sin ( ?) ?
? 20 t = ? – ?/3 = 2 ?/3 
? t = ?/30 sec.
c) v = A ? cos ( ?t + ?/3) = 20 × 5 cos (20t + ?/3)
when, v is maximum i.e. cos (20t + ?/3) = –1 = cos ?
? 20t = ? – ?/3 = 2 ?/3
? t = ?/30 sec. ?
7. a) x = 2.0 cos (50 ?t + tan
–1
0.75) = 2.0 cos (50 ?t + 0.643)
v = 
dt
dx
= – 100 sin (50 ?t + 0.643)
? sin (50 ?t + 0.643) = 0
As the particle comes to rest for the 1
st
time
? 50 ?t + 0.643 = ?
? t = 1.6 × 10
–2
sec.
Page 3


12.1
SOLUTIONS TO CONCEPTS 
CHAPTER 12
1. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w = 
T
2 ?
= 
6
2 ?
= 
3
?
sec
–1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ?) = 10 sin ? [y = r sin wt]
Sin ? = 1/2 ? ? = 
6
?
? Equation of displacement x = (10cm) sin ?
?
?
?
?
? ?
3
(ii) At t = 4 second
x = 10 sin 
?
?
?
?
?
? ?
? ?
?
6
4
3
= 10 sin 
?
?
?
?
?
? ? ? ?
6
8
= 10 sin ?
?
?
?
?
? ?
2
3
= 10 sin ?
?
?
?
?
? ?
? ?
2
= - 10 sin ?
?
?
?
?
? ?
2
= -10
Acceleration a = – w
2
x = –
?
?
?
?
?
?
?
?
?
9
2
× (–10) = 10.9 ? 0.11 cm/sec. ?
2. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec
–2
We know that a = ?
2
x 
? ? = 
x
a
= 
02 . 0
10
= 500 = 10 5
T = 
?
? 2
= 
5 10
2 ?
= 
236 . 2 10
14 . 3 2
?
?
= 0.28 seconds.
Again, amplitude r is given by v = ?
?
?
?
?
?
?
?
2 2
x r ?
? v
2
= ?
2
(r
2
– x
2
)
1 = 500 (r
2
– 0.0004)
? r = 0.0489 ? 0.049 m
? r = 4.9 cm. ?
3. r = 10cm
Because, K.E. = P.E.
So (1/2) m ?
2
(r
2
– y
2
) = (1/2) m ?
2
y
2
r
2
– y
2
= y
2
? 2y
2 
= r
2
? y = 
2
r
= 
2
10
= 5 2 cm form the mean position. ?
4. v
max
= 10 cm/sec.
? r ? = 10
? ?
2
= 
2
r
100
…(1)
A
max
= ?
2
r = 50 cm/sec
? ?
2
= 
y
50
= 
r
50
…(2)
Chapter 12
12.2
?
2
r
100
= 
r
50
? r = 2 cm.
? ? = 
2
r
100
= 5 sec
2
Again, to find out the positions where the speed is 8m/sec,
v
2
= ?
2
(r
2
– y
2
)
? 64 = 25 ( 4 – y
2
)
? 4 – y
2
= 
25
64
? y
2
= 1.44 ? y = 44 . 1 ? y = ?1.2 cm from mean position.
5. x = (2.0cm)sin [(100s
–1
) t + ( ?/6)]
m = 10g.
a) Amplitude = 2cm.
? = 100 sec
–1
? T = 
100
2 ?
= 
50
?
sec = 0.063 sec.
We know that T = 2 ?
k
m
? T
2
= 4 ?
2
×
k
m
? k = m
T
4
2
2
?
= 10
5
dyne/cm = 100 N/m. [because ? = 
T
2 ?
= 100 sec
–1
]
b) At t = 0
x = 2cm sin ?
?
?
?
?
? ?
6
= 2 × (1/2) = 1 cm. from the mean position.
We know that x = A sin ( ?t + ?)
v = A cos ( ?t + ?)
= 2 × 100 cos (0 + ?/6) = 200 × 
2
3
= 100 3 sec
–1
= 1.73m/s
c) a = – ?
2
x = 100
2
× 1 = 100 m/s
2
?
6. x = 5 sin (20t + ?/3)
a) Max. displacement from the mean position = Amplitude of the particle.
At the extreme position, the velocity becomes ‘0’.
? x = 5 = Amplitude.
? 5 = 5 sin (20t + ?/3)
sin (20t + ?/3) = 1 = sin ( ?/2)
? 20t + ?/3 = ?/2
? t = ?/120 sec., So at ?/120 sec it first comes to rest.
b) a = ?
2
x = ?
2
[5 sin (20t + ?/3)]
For a = 0, 5 sin (20t + ?/3) = 0 ? sin (20t + ?/3) = sin ( ?) ?
? 20 t = ? – ?/3 = 2 ?/3 
? t = ?/30 sec.
c) v = A ? cos ( ?t + ?/3) = 20 × 5 cos (20t + ?/3)
when, v is maximum i.e. cos (20t + ?/3) = –1 = cos ?
? 20t = ? – ?/3 = 2 ?/3
? t = ?/30 sec. ?
7. a) x = 2.0 cos (50 ?t + tan
–1
0.75) = 2.0 cos (50 ?t + 0.643)
v = 
dt
dx
= – 100 sin (50 ?t + 0.643)
? sin (50 ?t + 0.643) = 0
As the particle comes to rest for the 1
st
time
? 50 ?t + 0.643 = ?
? t = 1.6 × 10
–2
sec.
Chapter 12
12.3
b) Acceleration a = 
dt
dv
= – 100 ? × 50 ? cos (50 ?t + 0.643)
For maximum acceleration cos (50 ?t + 0.643) = – 1 cos ? (max) (so a is max)
? t = 1.6 × 10
–2
sec.
c) When the particle comes to rest for second time,
50 ?t + 0.643 = 2 ?
? t = 3.6 × 10
–2
s.
8. y
1
= 
2
r
, y
2
= r (for the two given position)
Now, y
1
= r sin ?t
1
?
2
r
= r sin ?t
1 
? sin ?t
1
= 
2
1
? ?t
1
= 
2
?
?
t
2 ?
× t
1
= 
6
?
? t
1
? ?
12
t
Again, y
2
= r sin ?t
2
? r = r sin ?t
2
? sin ?t
2
= 1 ? ?t
2
= ?/2 ? ?
?
?
?
?
? ?
t
2
t
2
= 
2
?
? t
2
= 
4
t
So, t
2
– t
1
= 
12
t
4
t
? = 
6
t
?
9. k = 0.1 N/m
T = 2 ?
k
m
?= 2 sec [Time period of pendulum of a clock = 2 sec]
So, 4 ?
2+ ?
?
?
?
?
?
k
m
= 4
? m = 2
k
?
= 
10
1 . 0
= 0.01kg ? 10 gm.
10. Time period of simple pendulum = 2 ?
g
1
Time period of spring is 2 ?
k
m
?
T
p
= T
s
[Frequency is same]
?
g
1
???
k
m
? ??
k
m
g
1
? ?
??????
k
mg
???
k
F
= x. (Because, restoring force = weight = F =mg)
? 1 = x (proved)
11. x = r = 0.1 m
T = 0.314 sec
m = 0.5 kg.
Total force exerted on the block = weight of the block + spring force.
T = 2 ?
k
m
?? 0.314 = 2 ?
k
5 . 0
? k = 200 N/m
? Force exerted by the spring on the block is 
F = kx = 201.1 × 0.1 = 20N
? Maximum force = F + weight = 20 + 5 = 25N
12. m = 2kg.
T = 4 sec.
T = 2 ?
k
m
? 4 = 2 ?
K
2
? 2 = ?
K
2
x
? ?
0.5kg
Page 4


12.1
SOLUTIONS TO CONCEPTS 
CHAPTER 12
1. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w = 
T
2 ?
= 
6
2 ?
= 
3
?
sec
–1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ?) = 10 sin ? [y = r sin wt]
Sin ? = 1/2 ? ? = 
6
?
? Equation of displacement x = (10cm) sin ?
?
?
?
?
? ?
3
(ii) At t = 4 second
x = 10 sin 
?
?
?
?
?
? ?
? ?
?
6
4
3
= 10 sin 
?
?
?
?
?
? ? ? ?
6
8
= 10 sin ?
?
?
?
?
? ?
2
3
= 10 sin ?
?
?
?
?
? ?
? ?
2
= - 10 sin ?
?
?
?
?
? ?
2
= -10
Acceleration a = – w
2
x = –
?
?
?
?
?
?
?
?
?
9
2
× (–10) = 10.9 ? 0.11 cm/sec. ?
2. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec
–2
We know that a = ?
2
x 
? ? = 
x
a
= 
02 . 0
10
= 500 = 10 5
T = 
?
? 2
= 
5 10
2 ?
= 
236 . 2 10
14 . 3 2
?
?
= 0.28 seconds.
Again, amplitude r is given by v = ?
?
?
?
?
?
?
?
2 2
x r ?
? v
2
= ?
2
(r
2
– x
2
)
1 = 500 (r
2
– 0.0004)
? r = 0.0489 ? 0.049 m
? r = 4.9 cm. ?
3. r = 10cm
Because, K.E. = P.E.
So (1/2) m ?
2
(r
2
– y
2
) = (1/2) m ?
2
y
2
r
2
– y
2
= y
2
? 2y
2 
= r
2
? y = 
2
r
= 
2
10
= 5 2 cm form the mean position. ?
4. v
max
= 10 cm/sec.
? r ? = 10
? ?
2
= 
2
r
100
…(1)
A
max
= ?
2
r = 50 cm/sec
? ?
2
= 
y
50
= 
r
50
…(2)
Chapter 12
12.2
?
2
r
100
= 
r
50
? r = 2 cm.
? ? = 
2
r
100
= 5 sec
2
Again, to find out the positions where the speed is 8m/sec,
v
2
= ?
2
(r
2
– y
2
)
? 64 = 25 ( 4 – y
2
)
? 4 – y
2
= 
25
64
? y
2
= 1.44 ? y = 44 . 1 ? y = ?1.2 cm from mean position.
5. x = (2.0cm)sin [(100s
–1
) t + ( ?/6)]
m = 10g.
a) Amplitude = 2cm.
? = 100 sec
–1
? T = 
100
2 ?
= 
50
?
sec = 0.063 sec.
We know that T = 2 ?
k
m
? T
2
= 4 ?
2
×
k
m
? k = m
T
4
2
2
?
= 10
5
dyne/cm = 100 N/m. [because ? = 
T
2 ?
= 100 sec
–1
]
b) At t = 0
x = 2cm sin ?
?
?
?
?
? ?
6
= 2 × (1/2) = 1 cm. from the mean position.
We know that x = A sin ( ?t + ?)
v = A cos ( ?t + ?)
= 2 × 100 cos (0 + ?/6) = 200 × 
2
3
= 100 3 sec
–1
= 1.73m/s
c) a = – ?
2
x = 100
2
× 1 = 100 m/s
2
?
6. x = 5 sin (20t + ?/3)
a) Max. displacement from the mean position = Amplitude of the particle.
At the extreme position, the velocity becomes ‘0’.
? x = 5 = Amplitude.
? 5 = 5 sin (20t + ?/3)
sin (20t + ?/3) = 1 = sin ( ?/2)
? 20t + ?/3 = ?/2
? t = ?/120 sec., So at ?/120 sec it first comes to rest.
b) a = ?
2
x = ?
2
[5 sin (20t + ?/3)]
For a = 0, 5 sin (20t + ?/3) = 0 ? sin (20t + ?/3) = sin ( ?) ?
? 20 t = ? – ?/3 = 2 ?/3 
? t = ?/30 sec.
c) v = A ? cos ( ?t + ?/3) = 20 × 5 cos (20t + ?/3)
when, v is maximum i.e. cos (20t + ?/3) = –1 = cos ?
? 20t = ? – ?/3 = 2 ?/3
? t = ?/30 sec. ?
7. a) x = 2.0 cos (50 ?t + tan
–1
0.75) = 2.0 cos (50 ?t + 0.643)
v = 
dt
dx
= – 100 sin (50 ?t + 0.643)
? sin (50 ?t + 0.643) = 0
As the particle comes to rest for the 1
st
time
? 50 ?t + 0.643 = ?
? t = 1.6 × 10
–2
sec.
Chapter 12
12.3
b) Acceleration a = 
dt
dv
= – 100 ? × 50 ? cos (50 ?t + 0.643)
For maximum acceleration cos (50 ?t + 0.643) = – 1 cos ? (max) (so a is max)
? t = 1.6 × 10
–2
sec.
c) When the particle comes to rest for second time,
50 ?t + 0.643 = 2 ?
? t = 3.6 × 10
–2
s.
8. y
1
= 
2
r
, y
2
= r (for the two given position)
Now, y
1
= r sin ?t
1
?
2
r
= r sin ?t
1 
? sin ?t
1
= 
2
1
? ?t
1
= 
2
?
?
t
2 ?
× t
1
= 
6
?
? t
1
? ?
12
t
Again, y
2
= r sin ?t
2
? r = r sin ?t
2
? sin ?t
2
= 1 ? ?t
2
= ?/2 ? ?
?
?
?
?
? ?
t
2
t
2
= 
2
?
? t
2
= 
4
t
So, t
2
– t
1
= 
12
t
4
t
? = 
6
t
?
9. k = 0.1 N/m
T = 2 ?
k
m
?= 2 sec [Time period of pendulum of a clock = 2 sec]
So, 4 ?
2+ ?
?
?
?
?
?
k
m
= 4
? m = 2
k
?
= 
10
1 . 0
= 0.01kg ? 10 gm.
10. Time period of simple pendulum = 2 ?
g
1
Time period of spring is 2 ?
k
m
?
T
p
= T
s
[Frequency is same]
?
g
1
???
k
m
? ??
k
m
g
1
? ?
??????
k
mg
???
k
F
= x. (Because, restoring force = weight = F =mg)
? 1 = x (proved)
11. x = r = 0.1 m
T = 0.314 sec
m = 0.5 kg.
Total force exerted on the block = weight of the block + spring force.
T = 2 ?
k
m
?? 0.314 = 2 ?
k
5 . 0
? k = 200 N/m
? Force exerted by the spring on the block is 
F = kx = 201.1 × 0.1 = 20N
? Maximum force = F + weight = 20 + 5 = 25N
12. m = 2kg.
T = 4 sec.
T = 2 ?
k
m
? 4 = 2 ?
K
2
? 2 = ?
K
2
x
? ?
0.5kg
Chapter 12
12.4
? 4 = ?
2
?
?
?
?
?
?
k
2
? k = 
4
2
2
?
? k = 
2
2
?
= 5 N/m 
But, we know that F = mg = kx
? x = 
k
mg
= 
5
10 2 ?
= 4
?Potential Energy = (1/2) k x
2
= (1/2) × 5 × 16 = 5 × 8 = 40J
13. x = 25cm = 0.25m
E = 5J
f = 5
So, T = 1/5sec.
Now P.E. = (1/2) kx
2
?(1/2) kx
2
= 5 ? (1/2) k (0.25)
2
= 5 ? k = 160 N/m.
Again, T = 2 ?
k
m
???
5
1
??????
160
m
? m = 0.16 kg.
14. a) From the free body diagram, 
? R + m ?
2
x – mg = 0 …(1)
Resultant force m ?
2
x = mg – R
? m ?
2
x = m ?
?
?
?
?
?
? m M
k
? x = 
m M
mkx
?
[ ? = ) m M /( k ? for spring mass system]
b) R = mg – m ?
2
x = mg - m x
m M
k
?
= mg –
m M
mkx
?
For R to be smallest, m ?
2
x should be max. i.e. x is maximum. ?
The particle should be at the high point.
c) We have R = mg – m ?
2
x 
The tow blocks may oscillates together in such a way that R is greater than 0. At limiting condition, R 
= 0, mg = m ?
2
x
X = 
2
m
mg
?
= 
mk
) m M ( mg ?
So, the maximum amplitude is = 
k
) m M ( g ?
15. a) At the equilibrium condition,
kx = (m
1
+ m
2
) g sin ?
? x = 
k
sin g ) m m (
2 1
? ?
b) x
1
= 
k
2
(m
1
+ m
2
) g sin ? (Given)
when the system is released, it will start to make SHM
where ? = 
2 1
m m
k
?
When the blocks lose contact, P = 0
So m
2
g sin ? = m
2 
x
2 
?
2
= m
2 
x
2
?
?
?
?
?
?
?
?
?
2 1
m m
k
? x
2
= k
sin g ) m m (
2 1
? ?
So the blocks will lose contact with each other when the springs attain its natural length.
A
B
M
K
x
m
mg
R
a= ?
2
x ?
m ?
2
x ?
m 2
x 2
m 1g
m 1
x 1
g
k
m 2g
F
R
(m 1 +m 2)g
R
P
m 2a
a
m 2g
Page 5


12.1
SOLUTIONS TO CONCEPTS 
CHAPTER 12
1. Given, r = 10cm.
At t = 0, x = 5 cm.
T = 6 sec.
So, w = 
T
2 ?
= 
6
2 ?
= 
3
?
sec
–1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ?) = 10 sin ? [y = r sin wt]
Sin ? = 1/2 ? ? = 
6
?
? Equation of displacement x = (10cm) sin ?
?
?
?
?
? ?
3
(ii) At t = 4 second
x = 10 sin 
?
?
?
?
?
? ?
? ?
?
6
4
3
= 10 sin 
?
?
?
?
?
? ? ? ?
6
8
= 10 sin ?
?
?
?
?
? ?
2
3
= 10 sin ?
?
?
?
?
? ?
? ?
2
= - 10 sin ?
?
?
?
?
? ?
2
= -10
Acceleration a = – w
2
x = –
?
?
?
?
?
?
?
?
?
9
2
× (–10) = 10.9 ? 0.11 cm/sec. ?
2. Given that, at a particular instant,
X = 2cm = 0.02m
V = 1 m/sec
A = 10 msec
–2
We know that a = ?
2
x 
? ? = 
x
a
= 
02 . 0
10
= 500 = 10 5
T = 
?
? 2
= 
5 10
2 ?
= 
236 . 2 10
14 . 3 2
?
?
= 0.28 seconds.
Again, amplitude r is given by v = ?
?
?
?
?
?
?
?
2 2
x r ?
? v
2
= ?
2
(r
2
– x
2
)
1 = 500 (r
2
– 0.0004)
? r = 0.0489 ? 0.049 m
? r = 4.9 cm. ?
3. r = 10cm
Because, K.E. = P.E.
So (1/2) m ?
2
(r
2
– y
2
) = (1/2) m ?
2
y
2
r
2
– y
2
= y
2
? 2y
2 
= r
2
? y = 
2
r
= 
2
10
= 5 2 cm form the mean position. ?
4. v
max
= 10 cm/sec.
? r ? = 10
? ?
2
= 
2
r
100
…(1)
A
max
= ?
2
r = 50 cm/sec
? ?
2
= 
y
50
= 
r
50
…(2)
Chapter 12
12.2
?
2
r
100
= 
r
50
? r = 2 cm.
? ? = 
2
r
100
= 5 sec
2
Again, to find out the positions where the speed is 8m/sec,
v
2
= ?
2
(r
2
– y
2
)
? 64 = 25 ( 4 – y
2
)
? 4 – y
2
= 
25
64
? y
2
= 1.44 ? y = 44 . 1 ? y = ?1.2 cm from mean position.
5. x = (2.0cm)sin [(100s
–1
) t + ( ?/6)]
m = 10g.
a) Amplitude = 2cm.
? = 100 sec
–1
? T = 
100
2 ?
= 
50
?
sec = 0.063 sec.
We know that T = 2 ?
k
m
? T
2
= 4 ?
2
×
k
m
? k = m
T
4
2
2
?
= 10
5
dyne/cm = 100 N/m. [because ? = 
T
2 ?
= 100 sec
–1
]
b) At t = 0
x = 2cm sin ?
?
?
?
?
? ?
6
= 2 × (1/2) = 1 cm. from the mean position.
We know that x = A sin ( ?t + ?)
v = A cos ( ?t + ?)
= 2 × 100 cos (0 + ?/6) = 200 × 
2
3
= 100 3 sec
–1
= 1.73m/s
c) a = – ?
2
x = 100
2
× 1 = 100 m/s
2
?
6. x = 5 sin (20t + ?/3)
a) Max. displacement from the mean position = Amplitude of the particle.
At the extreme position, the velocity becomes ‘0’.
? x = 5 = Amplitude.
? 5 = 5 sin (20t + ?/3)
sin (20t + ?/3) = 1 = sin ( ?/2)
? 20t + ?/3 = ?/2
? t = ?/120 sec., So at ?/120 sec it first comes to rest.
b) a = ?
2
x = ?
2
[5 sin (20t + ?/3)]
For a = 0, 5 sin (20t + ?/3) = 0 ? sin (20t + ?/3) = sin ( ?) ?
? 20 t = ? – ?/3 = 2 ?/3 
? t = ?/30 sec.
c) v = A ? cos ( ?t + ?/3) = 20 × 5 cos (20t + ?/3)
when, v is maximum i.e. cos (20t + ?/3) = –1 = cos ?
? 20t = ? – ?/3 = 2 ?/3
? t = ?/30 sec. ?
7. a) x = 2.0 cos (50 ?t + tan
–1
0.75) = 2.0 cos (50 ?t + 0.643)
v = 
dt
dx
= – 100 sin (50 ?t + 0.643)
? sin (50 ?t + 0.643) = 0
As the particle comes to rest for the 1
st
time
? 50 ?t + 0.643 = ?
? t = 1.6 × 10
–2
sec.
Chapter 12
12.3
b) Acceleration a = 
dt
dv
= – 100 ? × 50 ? cos (50 ?t + 0.643)
For maximum acceleration cos (50 ?t + 0.643) = – 1 cos ? (max) (so a is max)
? t = 1.6 × 10
–2
sec.
c) When the particle comes to rest for second time,
50 ?t + 0.643 = 2 ?
? t = 3.6 × 10
–2
s.
8. y
1
= 
2
r
, y
2
= r (for the two given position)
Now, y
1
= r sin ?t
1
?
2
r
= r sin ?t
1 
? sin ?t
1
= 
2
1
? ?t
1
= 
2
?
?
t
2 ?
× t
1
= 
6
?
? t
1
? ?
12
t
Again, y
2
= r sin ?t
2
? r = r sin ?t
2
? sin ?t
2
= 1 ? ?t
2
= ?/2 ? ?
?
?
?
?
? ?
t
2
t
2
= 
2
?
? t
2
= 
4
t
So, t
2
– t
1
= 
12
t
4
t
? = 
6
t
?
9. k = 0.1 N/m
T = 2 ?
k
m
?= 2 sec [Time period of pendulum of a clock = 2 sec]
So, 4 ?
2+ ?
?
?
?
?
?
k
m
= 4
? m = 2
k
?
= 
10
1 . 0
= 0.01kg ? 10 gm.
10. Time period of simple pendulum = 2 ?
g
1
Time period of spring is 2 ?
k
m
?
T
p
= T
s
[Frequency is same]
?
g
1
???
k
m
? ??
k
m
g
1
? ?
??????
k
mg
???
k
F
= x. (Because, restoring force = weight = F =mg)
? 1 = x (proved)
11. x = r = 0.1 m
T = 0.314 sec
m = 0.5 kg.
Total force exerted on the block = weight of the block + spring force.
T = 2 ?
k
m
?? 0.314 = 2 ?
k
5 . 0
? k = 200 N/m
? Force exerted by the spring on the block is 
F = kx = 201.1 × 0.1 = 20N
? Maximum force = F + weight = 20 + 5 = 25N
12. m = 2kg.
T = 4 sec.
T = 2 ?
k
m
? 4 = 2 ?
K
2
? 2 = ?
K
2
x
? ?
0.5kg
Chapter 12
12.4
? 4 = ?
2
?
?
?
?
?
?
k
2
? k = 
4
2
2
?
? k = 
2
2
?
= 5 N/m 
But, we know that F = mg = kx
? x = 
k
mg
= 
5
10 2 ?
= 4
?Potential Energy = (1/2) k x
2
= (1/2) × 5 × 16 = 5 × 8 = 40J
13. x = 25cm = 0.25m
E = 5J
f = 5
So, T = 1/5sec.
Now P.E. = (1/2) kx
2
?(1/2) kx
2
= 5 ? (1/2) k (0.25)
2
= 5 ? k = 160 N/m.
Again, T = 2 ?
k
m
???
5
1
??????
160
m
? m = 0.16 kg.
14. a) From the free body diagram, 
? R + m ?
2
x – mg = 0 …(1)
Resultant force m ?
2
x = mg – R
? m ?
2
x = m ?
?
?
?
?
?
? m M
k
? x = 
m M
mkx
?
[ ? = ) m M /( k ? for spring mass system]
b) R = mg – m ?
2
x = mg - m x
m M
k
?
= mg –
m M
mkx
?
For R to be smallest, m ?
2
x should be max. i.e. x is maximum. ?
The particle should be at the high point.
c) We have R = mg – m ?
2
x 
The tow blocks may oscillates together in such a way that R is greater than 0. At limiting condition, R 
= 0, mg = m ?
2
x
X = 
2
m
mg
?
= 
mk
) m M ( mg ?
So, the maximum amplitude is = 
k
) m M ( g ?
15. a) At the equilibrium condition,
kx = (m
1
+ m
2
) g sin ?
? x = 
k
sin g ) m m (
2 1
? ?
b) x
1
= 
k
2
(m
1
+ m
2
) g sin ? (Given)
when the system is released, it will start to make SHM
where ? = 
2 1
m m
k
?
When the blocks lose contact, P = 0
So m
2
g sin ? = m
2 
x
2 
?
2
= m
2 
x
2
?
?
?
?
?
?
?
?
?
2 1
m m
k
? x
2
= k
sin g ) m m (
2 1
? ?
So the blocks will lose contact with each other when the springs attain its natural length.
A
B
M
K
x
m
mg
R
a= ?
2
x ?
m ?
2
x ?
m 2
x 2
m 1g
m 1
x 1
g
k
m 2g
F
R
(m 1 +m 2)g
R
P
m 2a
a
m 2g
Chapter 12
12.5
c) Let the common speed attained by both the blocks be v.
1/2 (m
1
+ m
2
) v
2
– 0 = 1/2 k(x
1
+ x
2
)
2
– (m
1
+ m
2
) g sin ? (x + x
1
)
[ x + x
1
= total compression]
? (1/2) (m
1
+ m
2
) v
2
= [(1/2) k (3/k) (m
1
+ m
2
) g sin ? –(m
1
+ m
2
) g sin ???(x + x
1
)
? (1/2) (m
1
+ m
2
) v
2 
= (1/2) (m
1
+ m
2
) g sin ? × (3/k) (m
1
+ m
2
) g sin ?
? v = 
) m m ( k
3
2 1
?
g sin ?.
16. Given, k = 100 N/m, M = 1kg and F = 10 N
a) In the equilibrium position,
compression ??= F/k = 10/100 = 0.1 m = 10 cm
b) The blow imparts a speed of 2m/s to the block towards left.
?P.E. + K.E. = 1/2 k ?
2
+ 1/2 Mv
2
= (1/2) × 100 × (0.1)
2
+ (1/2) × 1 × 4 = 0.5 + 2 = 2.5 J
c) Time period = 2 ?
k
M
= 2 ?
100
1
= 
5
?
sec
d) Let the amplitude be ‘x’ which means the distance between the mean position and the extreme 
position.
So, in the extreme position, compression of the spring is (x + ?).
Since, in SHM, the total energy remains constant.
(1/2) k (x + ?)
2
= (1/2) k ?
2
+ (1/2) mv
2
+ Fx = 2.5 + 10x
[because (1/2) k ?
2
+ (1/2) mv
2
= 2.5]
So, 50(x + 0.1)
2
= 2.5 + 10x
? 50 x
2
+ 0.5 + 10x = 2.5 + 10x
?50x
2
=2 ? x
2 
=
50
2
= 
100
4
? x = 
10
2
m = 20cm.
e) Potential Energy at the left extreme is given by,
P.E. = (1/2) k (x + ?)
2
= (1/2) × 100 (0.1 +0.2)
2
=50 × 0.09 = 4.5J
f) Potential Energy at the right extreme is given by, 
P.E. = (1/2) k (x + ?)
2
– F(2x) [2x = distance between two extremes]
= 4.5 – 10(0.4) = 0.5J
The different values in (b) (e) and (f) do not violate law of conservation of energy as the work is done by 
the external force 10N.
17. a) Equivalent spring constant k = k
1
+ k
2
(parallel)
T = 2 ?
k
M
= 2 ?
2 1
k k
m
?
b) Let us, displace the block m towards left through displacement ‘x’ ?
Resultant force F = F
1
+ F
2
= (k
1
+ k
2
)x
Acceleration (F/m) = 
m
x ) k k (
2 1
?
Time period T = 2 ?
on Accelerati
nt displaceme
= 2 ?
m
) k k ( m
x
2 1
?
= 2 ?
2 1
k k
m
?
?
The equivalent spring constant k = k
1
+ k
2
c) In series conn equivalent spring constant be k. 
So, 
k
1
= 
1
k
1
+ 
2
k
1
= 
2 1
1 2
k k
k k ?
? k = 
2 1
2 1
k k
k k
?
T = 2 ?
k
M
= 2 ?
2 1
2 1
k k
) k k ( m ?
?
M
F
k
M
(a)
parallel
k 2
k 1
x-1
k 2
m
k 1  
k 2
m
k 1  
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