Download SIMPLE HARMONIC MOTION EXERCISE –I POLARISER

1
Problem File
SIMPLE HARMONIC MOTION
EXERCISE –I
POLARISER
1.
A simple harmonic motion is represented by F(t) = 10 sin (20t + 0.5). Write down its amplitude, anugular frequency,
frequency, time period and intial phase, if displacement is measured in m and time in s.
2.
A particle performs SHM between two points 0.2 m apart with a period 10 s. How far will the particle be in 1s, after
leaving the mid-position between two points.
3.
A particle is executing simple harmonic motion according to the equation x  5sin t where x is in cm. How long will
the particle take to move from the position of equilibrium to the position of maximum displacement.
4.
An object performs simple harmonic motion of amplitude 5 cm and period 4s. If timing is started when the object is
at the centre of an oscillation (i.e. x = 0), calculate
(a)
the frequency of the oscillation,
(b)
the displacement 0.5 s after the start.
(c)
the maximum acceleration of the system and
(d)
the velocity at a displacement of 3 cm
5.
A particle that is at the origin of co-ordinates at t = 0 vibrates about the origin along the y-axis with a frequency of
20 Hz and an amplitude of 3.0 cm. Given its equation of motion.
6.
A body executes 40 oscillations per minute. Its maximum speed is 36 cm s–1. Calculate the amplitude of oscillation.
7.
A particle executes a SHM of period  second and amplitude 2 cm. Find its acceleration when it is at the maximum
displacement from the mean position.
8.
The displacement x (in meters) of an oscillating particles varies with time t (in second) according to the


equation x  0.02cos 0.5t   . Calculate (a) amplitude of oscillation (b) time period of oscillation (c)
3


maximum velocity of particle (d) maximum acceleration of particle.
9.
A particle executes SHM of period 4 seconds and amplitude 0.02 m. Determine (i) its maximum speed, (ii) speed
at displacement 0.01 m, (iii) value of acceleration at the turning point, and (iv) value of acceleration at displacement
0.01 m.
10.
A particle executes SHM of amplitude 25 cm and time period 3s. What is the minimum time required for the particle
to move between two points 12.5 cm on either side of the mean position ?
11.
The vertical motion of a huge piston in a machine is approximately simple harmonic with a frequency of 0.50/s. A
block of 10 kg is placed on the piston. What is the maximum amplitude of the piston's SHM for the block and the
piston to remain together ?
12.
A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of
oscillation is 2.5×10–2 m. What must be the least period of these oscillations so that the objects is not detached
from the platform ? Take g = 10 m s–2.
13.
In a certain engine, a piston undergoes vertical SHM with amplitude 7 cm. A washer rests on top of the piston. As the
motor speed is slowly increased, at what frequency will the washer no longer stay in contact with the piston ?
14.
A particle is executing SHM with an amplitude of 4 cm and time period 12 second. The time taken by the particle in
going from its mean position to a position of displacement equal to 2 cm is t1. The time taken from this displaced
t1
position of 2 cm to reach the extreme position is t2. Calculate the ratio t .
2
15.
In what time after its motion begins, will a particle oscillating according to the equation x = 7 sin 0.5  t move from
the mean position to the maximum displacement ?
16.
A particle executing simple harmonic motion along a straight line has a speed of 4 m s–1 when at a distanne of 3 m
from the mean postion and 3 m s–1 when at a distance of 4 m from the mean position. Calculate the time it takes to
travel 2.5 m from the positive extremity of the oscillation. Take the value of  as 3 only..
17.
A small child bounces up and down while staying in contact with a trampoline. She undergoes SHM and repeats
her motion every 1.75 s. The height of each bounce above the equilibrium position of the trampoline is 35.0 cm.
(a) What is the amplitude and angular frequency of the child's motion ? (b) Write an equation of motion y(t) for the
child's motion if timing is to begin at the highest point of the motion. (c) What is the child's position at t = 4.50 s ?
18.
A particle of mass 0.2 kg execute simple harmonic motion of amplitude 2 cm and period 6 s. Find (i) total
mechanical energy at any instant (ii) kinetic and potential energies when the displacement is 1 cm.
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2
SIMPLE HARMONIC MOTION
19.
A metal cylinder of mass 4kg vibrates harmonically with an amplitude of 0.25 m and period of 2 second.
Calculate (i) the maximum acceleration and its acceleration when y = 0.15 m.(ii) its maximum restoring force
and restoring force when y = 0.15 m (iii) total energy.
20.
A particle executes SHM of amplitude a. (i) At what distance from the mean position is its kinetic energy equal to
potential energy (ii) At what points is its speed half the maximum speed.
21.
A 2kg body panel of a car oscillates with a frequency of 2 Hz and amplitude 2.5 cm. If the oscillations are assumed
to be simple harmonic and undamped, calculate
(a)
the maximum velocity of the panel
(b)
the total energy f the panel
(c)
the maximum potential energy of the panel, and
(d)
the kinetic energy of the panel 1.0 cm from its equlibrium position.
22.
A body of mass 9 kg is oscillating on a spring of force constant 100 N m –1. Deduce angular frequency.
23.
A massless spring steel is clamped at its lower end. A 2 kg ball is fastened to its top
end. A force of 8 N is required to displace the ball 20 cm to one side as shown.
Assume the system to undergo simple harmonic motion when released. Find (a)
the force constant of the spring and (b) the period with which the ball will vibrate
back and forth.
20 cm
8N
24.
A spring compressed by 0.1 m develops a restoring force of 10 N, A body of mass 4 kg is placed on it. Deduce (i)
the force constant of the spring (ii) the depression of the spring under the weight of the body (iii) the period of
oscillation if the body is disturbed. g = 10 m s–2.
25.
A spring of force constant 1200 N m –1 is mounted on a horizontal table as shown. A mass of 3.0 kg is attached to
the free end of the spring. The mass is attached to the free end of the spring. The mass is then pulled sideways to
a distance of 2.0 cm and released. What is
(a)
the speed of the mass when the spring is compressed by 1.0 cm.
(b)
potential energy of the mass when it momentarily comes to rest.
(c)
total energy of the oscillating mass ?
m
26.
A 3 g particles at the end of a spring moves according to the equation y = (0.75 sin 63t) cm where t is in second.
Find (a) the amplitude (b) frequency of motion (c) position at t = 0.020 s and (d) the spring constant.
27.
A 0.3 kg mass is connected to a system of two identical springs as shown in Fig. Determine the period of
oscillation.
k=20 Nm
–1
k=20 Nm
–1
0.3kg
28.
Fig shows the position function for a 0.500 kg mass executing SHM at the end of a spring. (a) What is the
amplitude, period and angular frequency for this SHM ? (b) What is the spring constant k ? (c) Find the position
x(t) for this mass. (d) What is the mass's velocity and acceleration at t = 5.00 s ?
8
6
4
2
0
–2
–4
–6
0
2
4
6
8
29.
What is the length of a pendulum for which time period is 1 s at a place where g = 0.8 m s–2 ?
30.
The time taken by a simple pendulum to perform 100 vibrations is 8 minute 9 second in Bombay and 8 minute 20
second in Pune. Calculate the ratio of accelerations due to gravity in Bombay and Pune.
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Problem File
31.
A vertical U-tube of uniform cross-section contains water upto a height of 2.45 vm. When the water on one side is
depressed and then released, its up and down motion in the tube is simple harmonic motion. Show that the time
period is equal to

seconds.
10
32.
A mass suspended by a spring oscillates in 100 cm3 of water. The force constant of the spring is 9800 N m–1. The
spring is stretched by 0.10 m and left to oscillate. Find the rise in temperature of water when the oscillations stop.
Assume that the whole of the energy of oscillation is used up in increasing the temperature of water.
33.
The simple harmonic motion of a particle is given by the equation y  3 sin t  4 cos t . Find its amplitude.
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SIMPLE HARMONIC MOTION
EXERCISE –II
10.
k1
(a)
k2
11.
m
3.
4.
5.
 k1  k 2 
(a) 2 

 m 
m
(b) 2 (k  k )
1
2
m(k1k 2 )
(c) 2 (k  k )
1
2
(d) 2
6.
7.
8.
m(k1  k 2 )
(k1k 2 )
12.

6sin  3t   and Y2 = 5(sin3  t+ 3 cos3  t).
4

Their amplitudes are in the ratio of :–
(a) 2 : 1
(b) 3 : 1 (c) 1 : 3 (d) 1 : 4
A mass M attached to a spring oscillates with a period
of 2 seconds. If the mass is increased by 2 kg the
period increases by 1 second. The initial mass M
will be :–
(a) 1.6 kg (b) 1 kg
(c) 1.5 kg (d) 2 kg
The ratio of kinetic enegy at mean position to the
potential enegy when the displacement is half of the
amplitude is :–
2
(b)
3
4
(c)
3
1
(d)
2
If the displacement (x) and velocity (v) of a particle
executing SHM are related through the expression
4v2 = 25 – x 2, then the time period is :–
(a) 
(b) 2 
(c) 4 
(d) 6 
In a simple pendulum at mean position :–
(a) K.E. is maximum and PE is minimum
(b) K.E. is minimum and PE is maximum
(c) Both P.E. and K.E. are maximum
(d) Both PE and KE are minimum
Maximum velocity in SHM is vm. The average velocity
during the motion from one extreme point to the other
extreme point will be :–
(a)

vm
2
(b)
2
vm

(c)
4
vm

(d)

vm
4
5
3
(b)
3
5
3
2
A
(b)
2A
3
A
4
(c)
(d)
25
9
(d)
3
A
2
Two blocks of mass m1 and m2 are kept on a smooth
horizontal table as shown in Fig. Block of mass m1
but not m2 is fastened to the spring. If now both the
blocks are pushed to the left so that the spring is
compressed a distance d. The amplitude of oscillation
of block of mass m1, after the system is released is :–
m1
13.
16
9
(c)
What will be the displacement of a particle in SHM
when its velocit is half the maximum velocity (A =
amplitude)
(a)
Two SHMs are represented by the equations Y1 =
4
(a)
1
When a particle oscillates simple harmonically,its
kinetic energy varies periodically. If frequency of
the particle is n, the frequency of the kinetic energy
is :–
(a) n/2
(b) n
(c) 2n
(d) 4n
A mass M is suspended from a spring of negligible
mass. The spring is pulled a little and then released,
so that the mass executes SHM of time period T. If
the mass is increased by m, the time period becomes
5T/3. The ratio of m/M is:–
m2
m1
(a) d m  m
1
2
m2
(b) d m  m
1
2
2m2
(c) d m  m
1
2
2m1
(d) d m  m
1
2
Displacement-time graph of a particle executing SHM
is shown. The correct corresponding force-time
graph of the particle is :–
y
Time
y
(a)
x
y
(b)
Force
9.
Force
2.
A simple harmonic motion (SHM) has an amplitude
A and time period T. The time required by it to travel
frm x = A to x = A/2 is :–
(a) T/6
(b) T/4
(c) T/3
(d) T/2
A mass m is suspended from two springs of spring
constant k1 and k2 as shown in the Fig. The time
period of vertical oscillations of the mass will :–
Force
1.
ANALYSER
x
Time
x
Time
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Problem File
(d)
x
x
Time
14.
Time
3E
and its displacement y is given by :–
4
15.
16.
(a)
A particle starts executing SHM of amplitude a and
total energy E. At the instant, its kinetic energy is
20.
a
a
a 3
(b) y 
(c) y 
(d)y = a
2
2
2
The periodic time of a mass suspended by a spring
(force constant k) is T. If the spring is cut in three
equal pieces, the force constant of each part and
the periodic time if the same mas is suspended from
one piece are :–
(a) y 
(a) k, T / 3
(b) 3k, T
(c) 3k, 3T
(d) 3k, T / 3
21.
22.
1
  v 2  v1  x 2  x1   2
(a) 

v 22  v12


23.
1
  v 2 x1 2   v1x 2 2  2

(b) 
v 22  v12


17.
1
m
2k
(b) 2
mgsin 
(c) 2
2k
18.
24.
On smooth inclined plane, a body of mass m is
attached between two massless springs. The other
ends of the springs are fixed to firm supports. If each
spring has force constant k, the period of oscillation
of the body is :–
(a) 2
(b)
2mgsin 
(d) 2
k
3
T
2

4
(c)
2
2
(d)
2
4
A particle is executing SHM with an amplitude of 4
cm. At the mean position, velocity of the particle is 10
cm/s. The distance of the particle from the mean
position when its speed becomes 5 cm/s is :–
(b)
5 cm (c) 2 3 cm
(d) 2 5 cm
The period of a simple pendulum is doubled, when :–
(a) Its length is doubled
(b) The mass of the bob is doubled
(c) Its length is made four times
(d) The mass of the bob and the length of the
pendulum are doubled
The period of oscillation of a simple pendulumof
constant length of earth surface is T. Its period inside
a mine is :–
(a) Greater than T
(b) Less than T
(c) Equal to T
(d) Cannot be compared
A simple pendulum has a time period T in vaccum.
Its time period when it is completely immersed in a
liquid of density one-eight of the density of material
of the bob is :–
7
8
3
8
T
T (c)
T (d)
T
(b)
8
5
8
7
A man measures the period of a simple pendulum
inside a stationary lift and finds it to be T s. IF the lift
accelerates upward with an acceleration g/4, then
the period of the pendulum will be :–
Pendulum
Lift
2m
k
A person measure the time period of a simple
pendulum inside a stationary lift and finds it to be T.
If the lift starts accelerating upwards with an
acceleration of g/3, the time period of the pendulum
will be :–
(a) 3 T
(b)
(a)
 v1x 2  v 2 x1  2
(d) 

2
2
 v1  v 2 
(c) v  v 2
 2 1

2
(a) 3 cm
x  A sin t , represents the equation of a SHM. If
displacement of the particle are x 1 and x 2 and
velocities are v1 and v2, respectively, then amplitude
of SHM is :–
v1x 2
The displacement x (in centimeters) of an oscillating
particle varies with time t (in seconds) as


x  2cos  0.5t   . The magnitude of the maximum
3


acceleration of the particle in cms–2 is :–
Force
Force
(c)
19.
y
y
2T
T
(c)
(d) 2T 5
5
4
What is the velocity of the bob of a simple pendulum
at its mean position. If it is able to rise to vertical
height of 10 cm (g = 9.8 m/s2) :–
(a) T
25.
(b)
A
(c) T / 3 (d) T/3
(a) 2.2 m/s
(c) 1.4 ms
B
(b) 1.8 m/s
M (d) 0.6 m/s
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26.
SIMPLE HARMONIC MOTION
The time period of a simple pendulum of length L as
measured in an elevator descending with acceleration
(a) 2
 3L 
(b)   
 g 
3L
g
 3L 
(c) 2  
 2g 
27.
(d) 2
K2
2L
3g
Two identical springs of constant K are connected
in series and parallel as shown in figure. A mass m
is suspended from them. The ratio of their
frequencies of vertical oscillations will be :–
31.
m
(B)
K
(a) 1  K 2
2
 1
1

(b) 

 2K1 K 2 
1
1
(c) 2K  K
1
2
2
1
(d)  

 K1 K 2 
(a)
(A)
(a) 2 : 1
(b) 1 : 1 (c) 1 : 2 (d) 4 : 1
As shown in figure, a simple harmonic motino oscillator
having identical four springs has time period :–
32.
k
k
k
33.
k
(a) T  2
(c) T  2
29.
m
4k
(b) T  2
m
k
(d) T  2
A mass m = 100 g is attached at the end of a light
spring which oscillates on a frictionless horizontal
table with an amplitude equal to 0.16 metre and time
period equal to 2 sec. Initially the mass is released
from rest at t = 0 and displacement x = –0.16 metre.
The expression for the displacement of the mass at
any time t is :–
(a) x = 0.16 cos( t )
30.
m
2k
2m
k

s
20
(b)

s
10
(c)

s
5
(d)

200
s
2  10 ) :–
(a) 6.3 m
(b) 0.63 m
(c) 0.25 m
(d) 63 cm
Two simple harmonic motions are represented by
(a) 7
34.
1
A body of mass 4.9 kg hangs from a spring and
oscillates with a period 0.5s. On the removal of the
body, the spring shortened by (Take g = 10 ms–2,

y1  4 sin(4t  ) and
2
resultant amplitude is :–
m
1
A weightless spring of length 60 cm and force constant
200 N/m is kept straight and unstretched on a smooth
horizontal table and its ends are rigidly fixed. A mass of
0.25 kg is attached at the middle of the spring and is
slightly displaced along the length. The time period of
the oscillation of the mass is :–
m
28.
K1
K1
g
is :–
3
(b) 1
y 2  3 cos(4t) .
(c) 5
The
(d) 2  3
Two simple harmonic motions are represented by
y1  5(sin2t  3 cos 2t)


y 2  5 sin  2t  
4

The ratio of the amplitude of two SHM's is :–
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1: 3
(b) x = –0.16cos( t )
(c) x = 0.16sin( t +  ) (d) x = –0.16sin(  t+  )
What will be the force constant of the spring system
shown in the figure :–
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Problem File
ANSWER KEY
Exercise–I
1.
10m, 20 rad
s–1 ,
3.18 Hz, 0.314 s, 0.5rad
s–2,
2. 58.8 mm
(d) ± 6.4
cms–1 5.
3.
4.
(a) 0.25 Hz, (b) 3.54 cm (c) –12.8 cm
y = (3.0 sin 125.6t) cm 6.
7.
8 cm s–2
9.
(i) 0.01 ms–1 , (ii) 2.72 ×10–2 m s–1 (iii) –49.4 × 10–3 ms–2 (iv) –0.0247 m s–2
10.
0.5 s
11. 0.99 m
12. 0.3142 s
14.
1
2
15. 1 s
16. 1 s
17.
(a) 35.0 cm, 3.59 cm s–1, (b) y(t) = 35.0 cos[(3.59 rad s–1)t] (c) –31.5 cm
18.
(i) 4.38 × 10–5 J (ii) 3.29×10–5 J, 1.10×10–5 J
19.
0.5 s
3.82 cm
8. (a) 0.02 m, (b) 4 s (c) 3.142×10–2 ms–1 (d) 4.94 × 10–2 m s–2
13.
above v = 1.88 Hz
(i) 0.252ms2 , 0.152ms2 (ii)  2 N,  0.6 2N , (iii) 0.125 2J
20.
(i)
21.
(a) 0.314 m s–1 , (ii) 0.1 J (c) 0.1 J, (d) 0.083 J
23.
(a) 40 N m–1 (b) 1.40 s
24.
(i) 100 N m–1, (ii) 0.4 m, (iii)
26.
(a) 0.75 m (b) 10.03 Hz, (c) 0.714 cm (d) 11.9 N m–1 27. 0.54 s
28.
 

1 
(a) 6.00 cm, 1.57 rad s–1 (b) 1.23 N m–1 (c) x(t) = (6.00cm) cos  rads  t  1.23rad (d) –3.24 cm s–1, – 14.0 cm s–2
2




29.
24.8 cm
33.
5 units
2
s
5
22. 3.33 rad s–1
a
2
(ii) ±
3
a
2
25. (a) 0.35 ms–1 (b) 0.24 j (c) 0.24 J
30. 1.0455
31.
32.
0.177ºC
Exercise–II
1.
A
2.
D
3.
D
4. A
5.
A
6. C
7.
A
8. B
9. C
10. C
11.
D
12.
A
13.
B
14. B
15. D
16. B
17. A
18. B
19. C
20. C
21.
C
22.
A
23. D
24. C
25. C
26. C
27. C
28. C
29. B
30. B
31.
A
32.
C
33.
34. C
A
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