Oscillation MCQ: Definition and Explanation

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The term oscillation refers to the continuous to and fro movement of an object between two positions or states. A particle is said to have SHM i.e. simple harmonic motion if it moves to and fro about a fixed point such that its acceleration at any instant is directly proportional to displacement but the direction of acceleration is opposite to the direction of displacement. Example: motion of spring on a frictionless surface when its one end is fixed and other is in motion.

Read Also: Periodic Motion

Ques- The path length of oscillating simple pendulum of length 1 m is 16 cm its max velocity is ( g = π2 m/s2 )

2 π cm/s
4 π cm/s
8 π cm/s
16 π cm/s

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Ans: (C)

Explanation:

(mgl – mgl cosθ) = ½ mv²

mgl ( 1 – cosθ) = ½ mv²

m is cancelled from both sides

2gl ( 1 – cosθ) = v²

2gl sin2θ = v² --------------------------------------- 1 – cosθ = sin²θ

v = √ (2 gl sin²θ)

Multiply and divide by 2

v = √ (4 g l sin²θ)/2

v = 2√ (g l sin²θ)/2

v = 2√ (g l sin²θ/4)

v = 2√ (g l) × sinθ/2

v = 2√ (g l) × θ/2 ------------------ sinθ/2 = θ/2

v = θ√ (g l)

We need to find θ

path length of oscillating simple pendulum

Ques- The distance covered by particle undergoing SHM in one time period is ------------- (Amplitude A)

4 A
Zero
A
2 A

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Ans: (A)

Explanation:

To understand the above problem, observed the figure carefully, in every half movement the particle covers distance A from its original position to right then come again its original position i.e. A +A = 2 A then pendulum goes left and covers distance A then come back its original position and covers distance A i.e. A+A =2A, it means total distance is 4 A

distance covered by particle undergoing SHM

Ques– The dimension of k in equation F = kx is ----------------

[ L⁰ M1 T^-2 ]
[ L² M1 T^-1 ]
[ L² M1 T^-2 ]
[ L⁰ M1 T^-1 ]

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Ans: (A)

Explanation:

F = Force - [ L¹ M¹ T^-2 ]

x – Distance – L¹

F / x = k

k = L⁰ M¹ T^-2

Ques- A particle performing SHM of periodic time ‘T’. The time taken by the particle in moving from means point to half the maximum displacement is ------------ ( Sin 30 = 0.5 )

T / 2
T / 4
T / 8
T / 12

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Ans: (D)

Explanation-

x = A sin ωt (from mean point)

x= A

âÂÂ---------------------->

x = A/2

A/2 = A sin ωt

½ = sin ωt

Sin-1 (1/2) = ωt

π/ 6 = 2π/T × t

t = T/12

Ques- Calculate the spring constant if the spring with 2N force is stretched by 40 cm

-5 N/m
0 N/m
8 N/m
-5 m/N

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Ans: (A)

Explanation-

F = -k x

F / x = -k

2N/0.4 m = -k

k = -5 N/ m

Ques- A particle is moving in a circle with uniform speed its motion is --------------

Periodic and simple harmonic
Periodic but not simple harmonic
Non periodic
None of the above

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Ans: (B)

Explanation-

If the speed of particle is uniform i.e. periodic in motion. But due to the retardation it will not be simple harmonic.

Ques- A small body of mass 0.10 kg is performing SHM of amplitude 1.0 m and period 0.20 S the maximum force acting on it is ...................

98.596 N
985.96 N
100.2 N
76.23 N

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Ans: (A)

Explanation-

F_max = m a_max ----------------- (1)

a_max = Aω²

a_max = 1× (2π / T)²

a_max = 1× (4π² / T²)

a_max = (4π² / 0.2²) ---------------------------- π = 3.14

a_max = 985.9 -------------------- put in (1)

F_max = 0.1 × 985.9

F_max = 98.59 N

Ques- The maximum velocity of particle in SHM is 0.16 m/s and maximum acceleration is 0.64 m/s2 . The amplitude is ------------------------

4×10^-2 m
4×10^-1 m
4×10 m
4×10⁰ m

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Ans: (A)

Explanation –

V_max = A ω --------------------------- (1)

a_max = A ω2 ---------------------------- (2)

Divide (2) by (1)

a_max/ V_max = Aω²/ Aω

ω = 0.64 / 0.16

ω = 4 ................... put in ( 1 )

0.16 = A ×4

A = 0.04

A = 4×10^-2 m

Ques- In a second pendulum mass of bob is 30 g if it is replaced by 90 g mass then its time period will be --------------------

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Ans: (B)

Explanation –

T = 2 π √ ( l / g)

The time period of second pendulum is always 2 sec

In this case you have to remember this condition.

Ques- The maximum velocity of a particle performing linear SHM is 0.16 m/s if its maximum acceleration is 0.64 m/s² . Calculate its period

2 s
1.7 s
1.5 s
1 s

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Ans: (C)

Explanation –

T = 2 π / ω

V_max = A ω --------------------------- (1)

a_max = A ω² ---------------------------- (2)

Divide (2) by (1)

a_max/ V_max = Aω²/ Aω

ω = 0.64 / 0.16

ω = 4

T = 2 π / 4

T = 1.5 s

Ques- Energy of a simple harmonic motion depends upon -----------

1 / ω²
ω
A²
1 / A²

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Ans: (C)

Explanation –

E. = ½ K A²
E. = ½ m ω² A²

From the above equation we can conclude that the energy of a simple harmonic motion depends upon the square of amplitude.

Ques- The SHM of a particle is given by the equation x = 3 sin (ωt + ωt) the amplitude is -------------

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Ans: (C)

Explanation –

To find the amplitude we need to focus on following equation

Displacement (x) = A sin (ωt + ωt)

A is amplitude

A = 3

Ques- If the velocity of a body is half the maximum velocity. Then what is the distance from the mean position?

2 A
√3 / 2 A
A
A / 2

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Ans: (B)

Explanation –

V = ω √(A² – x² )

V/2 = ω √(A² – x² )

Aω/2 = ω √(A² – x² )

A/2 = √(A² – x² )

Squaring both sides

A²/ 4 = (A² – x²)

A²/ 4 - A² = – x²

-3/4 A² = – x²

Negative sign will be canceled

√3 / 2 A = x

Physics Related Links:
Oscillatory Motion	Transverse and Longitudinal Waves
Wave Speed	Wave Power
Period Angular Frequency	Momentum
Frequency	Velocity
Angular Speed	Wave Motion

Oscillation MCQ: Definition and Explanation

CBSE CLASS XII Related Questions

1.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

4.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

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Oscillation MCQ: Definition and Explanation

CBSE CLASS XII Related Questions

1.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

2.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

4.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

5.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

2.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

3.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

4.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be: