Oscillations

QUEST TUTORIALS Head Office : E-16/289, Sector-8, Rohini, New Delhi, Ph. 65395439

Q1. The periodic beatings of the heart of a healthy person may be compared to a

(a) forced harmonic oscillator (b) damped harmonic oscillator

(c) forced harmonic oscillator with damping (d) pure free simple harmonic oscillator

Q2. Which one of the following functions of time represents a simple harmonic motion ?

(a) sin2 ωt + cos

2 ωt (b) e

−ωt + sin

2 ωt (c) sin ωt + cos 3 ωt (d) sin ωt + cos ωt

the symbols have usual meaning

Q3. The length of a seconds pendulum at a place where acceleration due to gravity is 10 m/s2 is about

(a) 1 m (b) 0.1 m (c) 10 m (d) 0.25 m

Q4. A particle executes S.H.M. of amplitudes 10 cm, the distance of a point from its mean position at which its kinetic

energy is exactly equal to its potential energy is about

(a) 0.71 cm (b) 7.1 cm (c) 71 cm (d) 0.51 cm

Q5. A simple pendulum mounted I a lift which acceleration downwards at 5 m/s2 at a place where g is 10 m/s

2, the

percentage change in the period is about

(a) 41% increase (b) 41% decrease (c) 20% decrease (d) 80% increase

Q6. The expression for the frequency of a damped harmonic oscillator with mass ‘m’ resting force constant ‘k’ and damping

factor ‘b’ is given by

(a) ( )2m2/bm/k − (b) ( )2

m2/bm/k2 −π (c)

2

m2

b

m

k

2

1

−

π (d)

2

m2

b

m

k

+

Q7. The most generated form of equation of motion of a forced simple harmonic oscillator with damping included is (where

the symbols have usual meaning)

(a) 0kxdt

xdm

2

2

=+ (b) tcosFkxdt

xdm 02

2

ω=+

(c) 0kxdt

dxb

dt

xdm

2

2

=++ (d) tcosFkxdt

dxb

dt

xdm 02

2

ω=++

Q8. In a simple harmonic motion which one of the following statements is incorrect ?

(a) the velocity leads the displacement by a phase of π/2

(b) the acceleration leads the velocity by a phase of π/2

(c) the displacement lags the acceleration by a phase of π/2

(d) the acceleration leads the displacement by a phase of π

Q9. The most general solution of 0kxdt

xdm

2

2

=+ representing a pure simple harmonic motion is of the form (where the

symbols have the usual meanings)

(a) A sin

t

m

k (b) A cos t

m

k

(c)

+ t

m

kcost

m

ksinA (d) A sin

φ+t

m

k

Q10. The speed (v) of a simple harmonic oscillator with amplitude A, in terms of its displacement (x) at any instant of time is

given by

(a) T

xAv

22 −= (b)

22xA

T

2v −

π= (c)

22xA

Tv −

π= (d)

22Ax

T

2v −

π=

Q11. The ratio of kinetic energy to the potential energy of a simple harmonic oscillator at a point mid way between the mean

equilibrium position and one of its extremities is

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

Q12. In simple harmonic motion which one of the following statements is true ?

(a) The magnitude of acceleration of a particle is the least at the extremities of the oscillator

(b) The total energy equals the potential energy at the end points together with the kinetic energy at the mean position

(c) the restoring force is maximum when the particle is instantaneously at rest during oscillations

(d) The time period primarily depends on the amplitude but independent of the phase


Q13. A spring balance has a scale that reads from O to 50 kg uniformly graduated on a scale of length 20 cm. The weight a

body suspended from this spring oscillating with a period of 0.63 s is about

(a) 25 Kg (b) 250 N (c) 2500 N (d) 250 Kg

Q14. An impulsive force gives an initial velocity of − 1.0 m/s to the mass attached to the free and of a spring which

subsequently oscillates with a time period of 0.63 s, the amplitude of acceleration is about

(a) 1 m/s2 (b) 10

−2 m/s

2 (c) 10

−1 m/s

2 (d) 10 m/s

2

Q15. In a S.H.M. the displacement (x) of mass ‘m’ as a function of time (t) is given by x = cos ,t6

−

π then its

(a) amplitude is 1 m and initial phase is − 30°

(b) amplitude is 1m and its time period is 1 s

(c) velocity amplitude is 1 m/s and initial phase is π/6 reds

(d) velocity amplitude is −1 m/s, and initial phase is −π/6 rad

Q16. The equalization due to gravity on the surface of moon is

(a) 1/6 of the acceleration due to gravity an earth

(b) 1/4 of the acceleration due to gravity an earth

(c) 1/3 of the acceleration due to gravity on earth

(d) 1/2 of the acceleration due to gravity

Q17. The percentage change in the time period of oscillation of a simple pendulum when it is taken from the surface of the

earth to the surface of moon, is about

(a) 145% (b) 14.5% (c) 145% increase (d) 145% decrease

Q18. The time period of oscillation of a seconds pendulum on the surface of moon is about

(a) 1.0 s (b) 2.0 s (c) 4.9 s (d) 2.5 s

Q19. If k1 and k2 are the spring constants of two springs arranged in parallel , their equivalent spring constant k (for k2 > k1)

is given by

(a) 21 k

1

k

1+ (b)

21 k

1

k

1− (c) k2 − k1 (d) k1 + k1

Q20. If k1 and k2 are force constants of two springs arranged in series their equivalent force constant (k) is given by

(for k2 > k2)

(a) k1 + k2 (b) 21 k

1

k

1+ (c)

21

21

kk

kk

+ (d)

21 k

1

k

1−

Q21. For a particle in linear S.H.M. the average kinetic energy over a period of oscillation is equal to

(a) total energy of the oscillator (b) half the total energy of the oscillator

(c) total kinetic energy of the oscillator (d) total potential energy of the oscillator

Q22. A spring of force constant k is cut into 3 equal parts. The force constant of each part is equal to

(a) k/3 (b) k (c) 3k (d) (2/3) k

Q23. Te time period of oscillation of a simple pendulum of length ''� is T, a plot of log T versus log ''� gives a straight line

with slope

(a) 2 (b) √2 (c) 1/√2 (d) ½

Q24. A simple pendulum has a hollow spherical bob filled with mercury. As the pendulum oscillates mercury gradually flows

out f the bob through a tiny hole at the bottom, the frequency of oscillation of the pendulum

(a) first decrease, then increases and finally attains its original value

(b) first increase, then decrease and finally attains its original value

(c) remain unaffected

(d) decreases continuously to a minimum

Q25. A pendulum is oscillating about an horizontal axis in a vertical plane in a lift. The acceleration with which the lift must

be raised to reduce the time period of oscillation to one half of its original value is oscillation to one half of its original

value is

(a) g/4 (b) 5g (c) 3g (d) 4g

Q26. A mass ‘m’ suspended from a light spring has a period T for its vertical simple harmonic oscillation. One adding a mass

‘M’ to ‘m’ the period becomes 3T, the mass added must be equal to

(a) 2m (b) 8m (c) 3m (d) 6m


Q27. The shape of the curve obtained from the plot of kinetic energy versus position of the bob of a simple pendulum from

its mean equilibrium position is a parabola opening

(a) upwards with its vertex at the origin

(b) upwards with its vertex anywhere on the x−axis

(c) downward with its vertex on the energy axis

(d) downwards with its vertex at the origin

Q28. In a S.H.M. the force (F) and the potential energy (U) are related to each other by

(a) ( ) ( );

dx

xdVxF = (b) ( ) ( )

dx

xdUxF −= (c) ( ) ( )

;dx

xudxF

2

2

−= (d) ( ) ( )∫−= xdvxF

Q29. The bob of a simple pendulum was allowed to oscillates simple harmonically by pulling the bob to the extreme right

and releasing it The period function exactly representing this S.H.M. is

(a) x = A sin ωt (b) x = A sin (ωt + φ) (c) x = A cos (ωt + φ) (d) x = A cos ωt

Q30. Combinations of two simple harmonic motions along two mutually perpendicular directions generates a special type of

figures called

(a) Fourier figures (b) Lissagou’s figures (c) Harmonic figures (d) circular figures

Q31. The frequency of oscillation of a critically damped ideal S.H.O. is

(a) very large (b) very small

(c) zero (d) equal to the frequency of undamped oscillator

Q32. A light spring has a force constant k1 and an object of mass m is suspended from it. The sprig is cut in half and the same

object is suspended from one of the halves. The ratio of frequencies of oscillations before and after the spring is cut is

equal to

(a) 2 (b) 1 (c) 1/√2 (d) √2

Q33. Any real spring has a mass of the mass of the spring (ms) of the simple harmonically oscillating spring−mass (m)

system is also taken into account, its period will

(a) increase (b) decrease

(c) remain unaffected (d) gets affected only if m >> ms

Q34. Damping devices are after used on machinery to

(a) increase the frequency of vibration (b) avoid resonance vibrations

(c) decrease the frequency of vibrations (d) critically damp the vibrations

Q35. In an electric shaver, the blade moves back and forth over distance of 2 mm. The motion is S.H.M. with a frequency

120 Hz. The velocity amplitude of the block is about

(a) 1.51 mm/s (b) 15.1 m/s (c) 1.51 m/s (d) 0.75 m/s

Q36. The piston in the cylinder heat of a locomotion has a stroke of 76.5 cm. The maximum speed of the piston when the

drive wheels make 193 rpm and the piston moves with S.H.M. will be

(a) 7.73 m/s (b) 15.46 m/s (c) 0.77 m/s (d) 20.20 m/s

Q37. A block is an a piston that is moving vertically with S.H.M. of period 1.18 s. The critical value of the amplitude of

motion at which the block just separate from the piston is (use g = 10 m/s2)

(a) 2.78 m (b) 278.0 m (c) 3.52 m (d) 35.2 cm

Q38. A loud speaker produces musical sound by the oscillation of a diaphragm. If the amplitude of oscillation is limited to ‘a’

metres the frequencies that would result in the acceleration of the diaphragm exceeding ‘g’, is

(a) greater than a/g2π (b) greater than ( ) a/g2/1 π

(c) less than a/g2π (d) less than a/g2

1

π

Q39. A body hung from a vertical light spiral spring stretches it by 1.8 cm. The frequency of oscillation body−spring system

is

(a) 3π/35 (b) 35π/3 (c) 35/3π (d) π/105

Q40. At a certain harbor, the tides cause the ocean surface to rise and toll in S.H.M., with a period of 12.5 hr. The time taken

for a tide to toll from its peak height to one−half its peck height above its average level is about

(a) 2.08 hr (b) 4.2 hr (c) 3.1 hr (d) 2.5 hr

Q41. A hypothetically large sling shot is stretched by 1.12 m to launch a 110 gm projectile with a speed sufficient to escape

from the earth (11.2 km/s) Assuming that an average person can exert a fore of 220 N, the number of persons required

to stretch the sling shot is


(a) 1.12 × 105 (b) 5.6 × 10

4 (c) 2.8 × 10

4 (d) 1.1 × 10

7

Q42. In a S.H.M. when the displacement is one half of the amplitude, the fraction of total energy which is kinetic and which

is potential are respectively equal to

(a) 0.25 and 0.75 (b) 0.50 and 0.50 (c) 0.75 and 0.25 (d) 0.67 and 0.33

Q43. In the forced S.H. oscillation of a damped block−spring system, if ω is the angular frequency then the ratio of amplitude

of oscillation to the maximum velocity is given by

(a) ω (b) 1/ω (c) 2π/ω (d) ω/2π

Q44. A particle executes S.H.M. with a frequency f. The frequency with which its kinetic energy oscillates is

(a) f (b) f/2 (c) 4f (d) 2f

Q45. When bullet in motion strikes a solid block of wood resting on a functionless table, and gets embedded into it,

(a) both the energy and momentum are conserved

(b) only momentum is conserved

(c) only energy is conserved

(d) neither momentum nor energy is conserved

Q46. A mass is hung from a vertical light spring and is set into oscillations. The frequency of oscillations can be decreased by

(a) taking the arrangement to a higher altitude

(b) increasing the amplitude of oscillation

(c) decreasing the mass hung

(d) adding identical spring to the first in series.


Answers

1. (c)

2. (d)

3. (a): Hint: T for a second’s pendulum is 2s, and T = 2π g/� , Find �

4. (b): Hint: the equation to be used is 1/(2) kx2 = 1/(2) k (A

2 − x

2), get x = a/√2

5. (a): Hint: (T′/T) = 'g/g and g′ = g −5, Find 100T

T'T×

− = 41%

6. (c): Hint: from theory ( ) ( )2m2/bm/k −=ω

7. (d)

8. (c)

9. (d)

10. (b) Hint: In S.H.M. we have 22

xAv −ω= s

11. (a): Hint: P.E. = 1/(2) kx2 = 1/(2) k (A/2)

2 = (1/4) E ; Hence K.E. = E

4

3

4

EE =−

12. (c)

13. (b): Hint: Mg = kx i.e. 50 g = 0.2 k ; T = k/M2π Find 2

2

4

kTgMg

π=

14. (d): Hint: accn. amp. = velocity amp. × ω

15. (c): Hint: compare with x = A cos (ωt + φ)

16. (a)

17. (c): Hint: meem g/gT/T = Find (Tm − Te) × 100/Te = 145%

18. (b)

19. (d): Hint: for springs in parallel keg = k1 + k2 + k3 +…

20. (c): Hint: for springs in series ...k

1

k

1

k

1

k

1

321eq

+++=

21. (b)

22. (c): Hint: for springs in series 321eq k

1

k

1

k

1

k

1++= Here k1 = k2 = k3 = k

23. (d): Hint: T = 2π g/� so log T = 1/(2) log � + log (2π/g)

24. (a): Hint: �/g2

1v

π= � increases and then decreases to initial value due to shift in the centre of gravity of the

pendulum.

25. (c): Hint: T′/T = 'g/g and g′ = 4g gives the result.

26. (b) Hint: T = 2π k/m

27. (c): Hint: K.E. = k/(2) (A2 − x

2)

28. (b) Hint: F(x) =

− 2kx

2

1

dx

d = −kx which S.H.O force

29. (d): Hint: at t = 0, x = +A and φ = 0 are given conditions

30. (b)

31. (c): Hint:

2

m2

b

m

k'

−=ω for km2b = ω′ = o critically damped.

32. (c): Hint: m/'k'w;m/kw == where 21

21

kk

kkk

+=

Here k1 = k2 = k′ ∴ k′ = 2k


33. (a): Hint: T = 2π ( ( ) k/3mm s+

34. (d)

35. (d): Hint: v0 = 2πvA; here A = 1 × 10−3

m & v = 120 Hz

36. (a): Hint: v = 2πvA; here A = 0.3825 m; v = 193/60 Hz

37. (a): Hint: |mg| = |kA| ; m/k = A/g Hence A = gT2/4π2

38. (b): Hint: a0 = Aω2 = g; hence A/g

2

1v

π=

39. (c): Hint: mg = kA; g/am/k ==ω

40. (a): Hint: x = a sin ωt hence 3

tT

2 π=

π and t = T/6 find x

41. (b): Hint: 1/(2) kA2 = 1/(2) mveq

2 find k to get kA/220 as the no of persons.

42. (c): Hint: P.E. = 1/(2) kx2 = 1/(4) ;

4

E

2

kA 2

=

and hence K.E. = 3E/(4)

43. (b): Hint: xm = Fm/bW ; vm = Fm/b and hence (xm/vm) = Yw

44. (d)

45. (b): Hint: mechanical energy is not conserved due to loss of energy

46. (d): Hint: T = 2π 2

kkk/m '

eq = for v′ < v.

Oscillations

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