Transcript Document

Physics
Session
Simple Harmonic Motion - 3
Session Objectives
Session Objective
Problems
Class Exercise - 1
O is a point of suspension of a simple
pendulum of length OA = 1.8 m. N is a
nail vertically below O, where ON = 0.9
m. The bob starts from A and returns to A
after a complete swing. What is the time
taken to complete the swing?
O
N
A
C
B
Solution
For the journey from A to B, let the time
be

T 1
t AB   2 
4 4 
g 
1
1.8 
 2
  0.67 s
4 
9.8 
Similarly, for tBC
'
2

Total time from A back to A is
1.8
 0.9
2
T' 1 
0.9 
tBC 
 2 
  0.47 s
4 4 
0.8 
t  2 t AB  t BC 
= 2[0.67 + 0.47]
= 2.28 s
Class Exercise - 2
Determine the period of small
oscillations of a mathematical pendulum,
i.e. a ball suspended by a thread = 20
cm in length, if it is located in a liquid
whose density is n = 3 times less than
that of the ball. The resistance of the
liquid is to be neglected.
Solution
In the figure, d and r are density of
ball and liquid respectively.
Then in equilibrium,
Upthrust = Vpg
T  V' rg  Vdg
T
or T  Vg(d  r)
Weight = Vdg
If A is the area and x is the
displacement of the liquid,
then ma  Axg(d  r)
Solution contd..
A g(d  r)
A g(d  r)
a
x 
x
m
A d
Compare this by equation a = w2x
So this represents the SHM. The time period T is given by
T
2

w
2
d
 2
g(d  r)
g(d  r)
d
Also d = 3r
T = 1.1 s
Class Exercise - 3
A block is resting on a piston which
is moving vertically with an SHM of
period 1 s. At what amplitude of
motion will the block and piston
separate?
Solution
For the block,
mg – R = ma
or R = m(g – a)
In order to separate the block, R = 0 or a = g
R
Now a = w2x
 g = w2x
or x 
mg
g
w2

9.8
 
2
T
2

9.8
4 2
= 0.248 m
Class Exercise - 4
A plank with a body of mass m placed
on it starts moving straight up
according to the law X = a(1 – coswt),
where X is the displacement from the
initial position, w = 11 s–1. Find the time
dependence of the force that the body
exerts on the plank. If a = 4.0 cm plot
this dependence.
Solution
X  a 1  cos w t 
dx
d2x

 aw sin wt and
 aw2 cos wt
dt
dt 2
Now F  m
d2x
dt 2
 mg
 ma w2 cos w t  mg


a w2
 mg  1 
cos w t 


g


Plot is shown in the figure.
1.5
1.0
0.5
O

2
wt
Class Exercise - 5
Find the time period of a pendulum
of infinite length. Assume bob to be
near to surface of earth.
O

T
l
B
A
mg
R 
O´
  
Solution
Restoring force on the bob = mg sin( + )
Force on the bob = –mg{sin( + )}
Acceleration a 
force
 gsin     
mass
If  is small,  is also small.
Hence, sin(  )    
a   g      ; also  
AB
and  
AB
R
Solution contd..
 AB AB 
a  g 

R 

R  
 g
AB

 R 
R  
So w2  g 

 R 
R
t  2
(R  )g
R
t  2
g
[If
is infinite]
Class Exercise - 6
In the arrangement shown in the figure,
the particle m1 rotates in a radius r on a
smooth horizontal surface with angular
velocity w0. If m2 is displaced slightly in
the vertical direction, find the time
period of oscillation.
wo
O r
m2
m1
Solution
Considering the equilibrium of m2, we have
m1rw02  m2g
w0 
m 2g
m1r
Now let mass m2 is displaced downward by a
distance x. Then radius of circular path of m1
decreases to (r – x). Applying the conservation
of angular momentum, we have
m1r 2w0  m1(r  x)2  w
Solution contd..
 r 
or w  w0 

r

x


2
Tension is also increased as shown below.
T  m1w2
r4
(r  x)3

mw0 2r
x

1



r


3
 3x
 m1w0 2r 1 
r 

As a result, m2 gets a restoring force given by
F   T  m2g


 3x 
2
or F   m1w0 r 1 
  m2g
r 



Solution contd..
or F  3m1w02x

or m1  m2
or
d2 x
dt2

d2x
 dt 2
 3m1w0 2x
3m1w0 2
m1  m 2

x


 m1  m2 r 

So T  2  

3m
g


2


Class Exercise - 7
Find the frequency of small oscillations
of a thin uniform vertical rod of mass
m and length
is hinged at point O.
The combined stiffness of the spring is
equal to K. The mass of the spring is
negligible.
O
l
Solution
A displaced position of the rod through
an angle  is shown in the given
figure. The displacement of spring is x.
Let K1 and K2 be the stiffness of the
springs respectively.
O

Considering the torques acting
on the rod, we have
m
g
(K1+K)
2x
K 1  K 2  x
cos   mg sin   I
2
Solution contd..
When  is small, cos  = 1 and sin  = 
m 2
 K1  K 2 x  mg  

2
3
x
Now




So K1  K 2

2
2  mg   m

2
3
Solving it for , we get

 3m g  
 3 K1  K 2   2  

    w 2

m






Hence, the motion is SHM.


So w 

 3m g  
3K




2



m






Class Exercise - 8
In the arrangement shown in the
following figure, the sleeve M of
mass m = 0.2 kg is fixed between
two identical spring whose combined
force constant K = 20 N/m. The
sleeve can slide without friction over a horizontal
bar AB. The arrangement rotates with a constant
angular velocity w = 4 rad/s about a vertical axis
passing through the middle of the bar. Find the
period of small oscillations of the sleeve?
w
Solution
We will analyse the problem relative to
the rotating bar AB. As the acceleration
of bar will be centripetal, a pseudo force
will act on the sleeve away from centre
and will be of magnitude m w2x.
A
x
B
m
If the sleeve is displaced by
x1, the net force towards the
centre is


F  K x  m w2x  K  m w2 x
w
 Period  2 
m
K  m w2
Solution contd..
Note: If K  m w2, there will be no
oscillation of the sleeve. It will rush to
the point B if it is displaced slightly
(for K < mw2) or will remain in the
displaced position (for K = mw2)
Class Exercise - 9
The friction coefficient between the
two blocks shown in the figure is m
and the horizontal plane is smooth.
(a) Find the time period of system, and
(b) magnitude of frictional force between the blocks.
m
M
Solution
(a) For small amplitude, the two blocks
oscillate together. In this case,
w
mM
K
or T  2
mM
K
(b) The acceleration of the blocks at displacement
x from mean position
Kx
2
a  w x 
Mm
Force on upper block = ma = 
mKx
Mm
Solution contd..
The force is provided by friction of the
lower block.
 Magnitude of frictional force =
mK x
Mm
Class Exercise - 10
A liquid of density d is kept in a vertical
U-tube of uniform cross section A. If
the liquid column is slightly depressed
and left, show that the resulting motion
of the liquid is SHM and find the period.
Solution
Let the liquid be depressed by a
height x in the right side of the Utube. Then the liquid rises above on
the left-side by a height x.
 Restoring force = Pressure × Area
x
= – (2xdg) × A
F  x
or K = 2dgA
or T  2 
m
m
 2
K
2dgA
 The motion is SHM.
Thank you