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

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

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.

If x is the displacement then,

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

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

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

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

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.

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.

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.

Case (i) : Applying conservation of linear momentum.
 


Clearly E₂ < E₁
The new time Period 
Case (ii) : The new time Period 

Also A₂ = A₁
Here E₂ = E₁
The instantaneous value of speed at X_o of the combined masses decreases in both the cases.