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Test: MCQs (One or More Correct Option): Simple Harmonic Motion (Oscillations) | JEE Advanced

11 Questions MCQ Test Physics 35 Years JEE Main & Advanced Past year Papers | Test: MCQs (One or More Correct Option): Simple Harmonic Motion (Oscillations) | JEE Advanced

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Attempt Test: MCQs (One or More Correct Option): Simple Harmonic Motion (Oscillations) | JEE Advanced | 11 questions in 20 minutes | Mock test for JEE preparation | Free important questions MCQ to study Physics 35 Years JEE Main & Advanced Past year Papers for JEE Exam | Download free PDF with solutions

*Multiple options can be correct

QUESTION: 1

A particle executes simple harmonic motion with a frequency. f. The frequency with which its kinetic energy oscillates is

A.
f/2
B.
f
C.
2f
D.
4f

Solution:

NOTE : During one complete oscillation, the kinetic energy will become maximum twice.
Therefore the frequency of kinetic energy will be 2f.

*Multiple options can be correct

QUESTION: 2

A linear harmonic oscillator of force constant 2 × 10⁶ N/m and amplitude 0.01 m has a total mechanical energy of 160 J.
Its

A.
maximum potential energy is 100 J
B.
maximum kinetic energy is 100 J
C.
maximum potential energy is 160 J
D.
maximum potential energy is zero

Solution:

The total energy of the oscillator

As total mechanical energy = 160 J
The P.E. at equilibrium position is not zero.
P.E. at mean position = (160 – 100) J = 60 J
∴ Max P.E. = (100 + 60 ) J = 160 J.

*Multiple options can be correct

QUESTION: 3

A uniform cylinder of length L and mass M having cross sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half- submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is

Solution:

If x is the displacement then,

*Multiple options can be correct

QUESTION: 4

A highly rigid cubical block A of small mass M and side L is fixed rigidly on to another cubical block B of the same dimensions and of low modulus of rigidity η such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the sides faces of A. After the force is withdrawn, block A executes small oscillations the time period of which is given by

Solution:

NOTE : When a force is applied on cubical block A in the horizontal direction then the lower block B will get distorted as shown by dotted lines and A will attain a new position (without distortion as A is a rigid body) as shown by dotted lines.

For cubical block B

⇒ F = ηLΔL

ηL is a constant

⇒ Force F ∝ ΔL and directed towards the mean position, oscillation will be simple harmonic in nature.
Here, Mω² = ηL

*Multiple options can be correct

QUESTION: 5

One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K.A mass m hangs freely from the free end of the spring. The area of cross-section and the Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to:

A.
2π(m /K)^1/2
B.
C.
2π[(mYA / KL)^1/2
D.
2π[(mL / YA)^1/2

Solution:

Let us consider the wire also as a spring. Then the case becomes that of two spring attached in series. The equivalent spring constant is

where k' is the spring constant of the wire

We know that time period of the system

*Multiple options can be correct

QUESTION: 6

A particle of mass m is executing oscillations about the origin on the x axis. Its potential energy is V(x) = k | x |³ where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

A.
proportional to 1/ √a
B.
independent of a
C.
proportional to √a
D.
proportional to a^3/2

Solution:

U (x) = k | x |³

Now time period may depend on T ∝ (mass)^x (amplitude)y (k)^z
∴ [M⁰L⁰T] = [M]^x [L]^y [ML^–1 T^–2]z
= [M^{x + z} L^{y – z} T ^{– 2z}]
Equating the powers, we get
– 2z = 1 or z = – 1/2
y – z = 0 or y = z = – 1/2
Hence T ∝ (amplitude) ^–1/2 ∝ a^–1/2

*Multiple options can be correct

QUESTION: 7

Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by 45°, then.

A.
the resultant amplitude is (1 + √2)a
B.
the phase of the resultant motion relative to the first is 90°
C.
the energy associated with th e resulting motion is (3 + 2√2) times the energy associated with any single motion
D.
the resulting motion is not simple harmonic.

Solution:

From superposition principle

y = y₁ + y₂ + y₃

= a sin ωt + a sin (ωt + 45°) + a sin (ωt + 90°)
= a[sin ωt + sin (ωt + 90°] + a sin (ωt + 45°)
= 2a sin (ωt + 45°) cos 45° + a sin (ωt + 45°)

= (√2 +1) a sin (ωt + 45°)
= A sin (ωt + 45°)

Therefore resultant motion is simple harmonic of amplitude A = (√2 +1) a

and which differ in phase by 45° relative to the first.
Energy in SHM ∝ (amplitude)²

*Multiple options can be correct

QUESTION: 8

The function x = A sin² ωt + B cos² ωt + C sin ωt cosωt represent SHM for which of the option(s)

A.
for all value of A, B and C (C ≠ 0)
B.
A = B, C = 2B
C.
A = -B, C = 2B
D.
A = B, C = 0

Solution:

The given equation is

x = A sin² ωt + B cos² ωt + C sin ωt cos ωt

NOTE THIS STEP

Rearranging the equation in a meaningful form (for interpretation of SHM)

The above equation is that of SHM with amplitude C/2 and angular frequency 2ω. Thus option (a) is correct.

(b) If A = B and C = 2B then x = B + B sin 2ωt
This is equation of SHM. The mean position of the particle executing SHM is not at the origin.
Option (b) is correct.
(c) A = – B, C = 2B; Therefore
x = B cos 2ωt + B sin 2ωt
Let B = X cos φ = X sin φ then
x = X sin 2ωt cos φ + X cos 2ωt sin φ This represents equation of SHM.
(d) A = B, C = 0 and x = A.
This equation does not represents SHM.

*Multiple options can be correct

QUESTION: 9

A metal rod of length ‘L’ and mass ‘m’ is pivoted at one end.
A thin disc of mass ‘M’ and radius ‘R’ (<L) is attached at its center to the free end of the rod. Consider two ways the disc is attached: (case A). The disc is not free to rotate about its centre and (case B) the disc is free to rotate about its centre.
The rod disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true?

A.
Restoring torque in case A = Restor in g torque in case B
B.
restoring torque in case A < Restoring torque in case B
C.
Angular frequency for case A > angular frequency for case B.
D.
Angular frequency for case A < Angular frequency for case B.

Solution:

Applying τ = Iα

The restoring torque in both the cases is same.
Also I_A > I_B ∴ α_A < α_B

∴ ω_A < w_B

(a) and (d) are correct options.

*Multiple options can be correct

QUESTION: 10

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω₁ and ω₂ and have total energies E₁ and E₂, respectively. The variations of their momenta p with positions x are shown in the figures. then the correct equation(s) is(are)

A.
E₁ω₁ = E₂ω₂
B.
C.
ω₁ω₂ = n²
D.

Solution:

Maximum linear momentum in case 1 is (p₁)_max = mv_max
b = m [aw₁] ...(i)
Maximum linear momentum in case 2 is (p₂)max = mv_max
R = m [Rω₂]

∴ 1 = mω₂ ...(ii)

Dividing (i) & (ii)

∴ B is a correct option.

*Multiple options can be correct

QUESTION: 11

A block with mass M is conn ected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x₀. Consider two cases: (i) when the block is at x₀; and (ii) when the block is at x = x₀ + A. In both the cases, a particle with mass m(<M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?

A.
The amplitude of oscillation in the first case chan ges by a factor of whereas in the second case it remains unchanged.
B.
The final time period of oscillation in both the cases is same.
C.
The total energy decreases in both the cases.
D.
The instantaneous speed at x₀ of the combined masses decreases in both the cases