Physics mcqs on periodic motion for competitive exams preparation part 2

21. What is the number of degrees of freedom of an oscillating simple  pendulum?

(a) 1               (b) 2
(c) 3               (d) 4

 

There are 2 degrees of freedom, 1 translational along which the pendulum bob moves, and second rotational - the hinge about which it forms an arc motion.

 

 

 

22. The graph between restoring force and time in SHM is

(a) straight line     (b) parabola
(c) sine curve         (d) circle

 

 

sine curve

We know that for SHM, x=Asin(ωt+θ) and F=kx=kAsin(ωt+θ)
Thus it's a sine curve.

 

 

 

23. How much radians is the phase difference between the velocity and displacement is SHM?

(a) π                     (b) π/ 2

(c) 2π

(d) 0

 

π/2 radian

x=Asin(ωt+θ)
v=dx/dt= Aωcos(ωt+θ)=−Aωsin(ωt+θ+π/2) since cos(π/2+x)=−sinx 
Thus, phase difference is of π/2

 

 

 

24. What is the time period of a seconds pendulum?

 

a)    1 sec          b)   2 sec

c)    3 sec          d)   4 sec

 

 

 

 

 

 

 

25. The time period of the hour hand of a watch is

 

(a) 1 h                 (b) 6 h
(c) 12 h               (d) 24 h

12 h

Hour hand completes 1 rotation, 3600 or 2π in 12 hours.
Thus, T=12 hours

 

 

 

   

26. The curve between the acceleration and velocity of body in SHM is a (an)

(a) circle             (b) parabola
(c) ellipse           (d) triangle

 

ellipse

We have x=Asin(ωt)
Thus, v=dx/dt=Aωcos(ωt)
Also, a=−ω2x=−Aω2sin(ωt)
Thus, we have sin2(ωt)+cos2(ωt)=1,    giving (−a​/ Aω2)2+(v/Aω​)2=1
or, v2ω2+a2=A2ω4=c2, which is the equation of an ellipse

 

 

 

27. Which of the following exhibits chaotic behaviour?

 

(a) double pendulum    (b) inverted pendulum
(c) pendulum                   (d) none of the above

 

What is chaotic behavior?

The mathematical conception that some phenomena that seem random may be of a deterministic order highly sensitive to initial conditions and perturbations

 

28. The potential energy of a simple pendulum at rest is 10 J and its mean kinetic energy is 5 J. Its total energy at any instant will be

 

 (a) 5 J            (b) 10 J
(c) 15 J           (d) 20 J

 

10 J

The total energy of the system remains constant. Since it is given that P.E at rest is 10J, the total energy must be 10J as K.E at rest is 0. The mean K.E is useless data. Hence option B is correct.

 

 

 

 

29. The time period of a torsional pendulum is

(a) T = π          (b) T = 2π 

(c) T = 2π       (d) T = π                      

       T= 2π

In torsional pendulum, ω=
T=2π/ ω ​= 2π
Option C is correct.

 

 

30. Restoring force in the simple harmonic motion is

(a) centripetal               (b) frictional
(c) conservative           (d) non conservative

conservative

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Restoring force F=kx, is such kind of force and is conservative.

 

 

31. The time period of the second’s hand of a watch is

 

(a) 1 s              (b) 1 min            

(c) 1 h             (d) 12 h

 

1 min

Time taken by second's hand to complete one full rotation of 360 degrees is 1 minute, which is its time period .

 

 

 

32.  The mean kinetic energy of a harmonic oscillator with to position is

 

(a)  Ka2 /2        b)  Ka2 /3       

 c)   Ka2 /4       d)  Ka2 /6  

 

ka2/3​

 

The average K.E can be given by 
K.E=​​k(a2−x2)= ka3​/3
Hence, average KE with respect to mean position will be 
​​= 
Option C is correct.

 

 

    

33. What is the maximum time period of a simple pendulum?

(a) 84.6 min        (b) 1 day
(c) 12 h                (d) 1 year

 

 

T=2π              On the surface of the earth if we suppose a pendulum thread has a length of 6400000 m                 (=Radius of the Earth) (completely hypothetical though) we get the   time period as :                       

                                                                                                                  

=2π                          The situation presented here is completely hypothetical becuse of :

=5077.581s                                                 1. Magnitude of gravitational acceleration decreases as we move away from the
                                                                                                                  surface of the earth so the value of g for the whole pendulum is not constant
                                                                                                                 at all points with maximum at surface of Earth (on the equator) = 9.86 m/s
                                                                                                                 and the lowest at the pendulum bob = 2.45 m/s .
                                                                                                            2. A pendulum with 6400 km as length is quite unimaginable and is quit

                                                                                                               impossible to set up.

=84.6min

34. The differential equation representing SHM of a particle is d2y /dt2 + ω2y = 0

(a) ω                       (b) ω/ π
(c)  ω/2π              (d) 2πω

 

 

 

35. The equation of a harmonic oscillator is given by d2y /dt2 + ky = 0, where k is a positive constant. What is the time period of motion?

(a) 2π/k           (b) 2πk

(c) 2π/                d) 2 π                

d2y/dt2   = -ky                           . d2y/dt2   is acceleration

so,  a= -ky = -w2y                

w2 = k   , w=                           .T=2π/w

so,   T=2π/

 

 

36. The displacement equation of a harmonic oscillator is given by y = Asin ωt - Bcost ωt the amplitude of the particle is

(a) A + B                  (b) A - B
(c) A2 + B2               (d)

 

 

Displacement equation
y=Asinωt−Bcosωt
Let A=acosθ and B=asinθ
So, A2+B2=a2
⇒ a=
Then y=acosθsinωt−asinθcosωt
y=asin(ωt−θ)
which is the equation of simple harmonic oscillator
The amplitude of the oscillator
a=

 

37. A spring of force constant k is cut into three equal parts. The force constant of earth part will be

(a) k                      (b) 3k

  c) k/3                  d) zero

 

3K
The force constant is inversely proportional to the length of the spring.

F=kx

k=F/x​, where x is the length of the spring

Since, the spring has been divided into three parts, so the length of each part becomes one third of the length. Thus, the spring constant of each part would be three times the spring constant of the complete spring.

 

 

 

38. A particle is executing SHM with an amplitude 4 cm. At what displacement its energy is half kinetic and half potential?

(a) 2 cm                 (b) 1 cm
(c)   cm        (d) 2  cm

 

 ANS:   2​cm

In SHM the KE and PE of particle is given by the following relations:

KE=1/2mω2(a2−x2)

PE=1/2​mω2x2

So, the total energy,

TE=KE+PE

TE=1/2 mω2a2

As, PE = KE (both half). From above equations:

a2−x2=x2

2x2=a2

x=a/​          

x=​4​ /​                                               AS  a=4​cm

x=2​     cm     

 

 

39. A particle executing SHM has velocity 4 cm/s and 3 cm/s when its distances from mean positions is 2 cm and 3 cm respectively. The period of oscillation of the particle is (in seconds)

 

a) 5.3 s                        b) 2.94 s

c)  7.31 s                   d) 9.31 s

 

ANS :  5.3s

Velocity of particle executing SHM at 2cm  from mean position v1=4cm/s and at 3cm from mean
position . v2 = 3 cm/s
v₁ = ω  = ω  
v₂ = ω   = ω                         .v=rw

=    =

16a2-144 =9a2 -36

a2 =

Putting the value of in first equation we get :,

ω = 1.18 ms-1
 Time period T =       = 5.3 s

 

40. A simple pendulum suspended from the ceiling form a lift has a period T, when the lift falls freely, the time period of pendulum will become

(a) zero             (b)  T /9.8
(c) 9.8 T            (d) infinity

 

Time period of simple pendulum accelerating downward is given by,

T=2

where,
l= length of the pendulum

g= acceleration due to gravity

  a=   acceleration of the pendulum
But, when the lift falls freely under gravity, the net acceleration acting on the pendulum will be zero.

f=0    ⟹  T’ = ∞
Since the frequency of oscillation is reciprocal of the time period. So,

⟹ frequency = 0

When the lift falls freely, effectively acceleration and frequency of oscillation becomes zero.

 

ANSWERS

21. b 22. c 23. b 24. b
25. c 26. c 27. a 28. b
29. c 30. c 31. b 32. b
33. a 34. c 35. c 36. d
37. c 38. d 39. a 40. D

 

<<BACK