Test: Simple Harmonic Motion(9 Oct) - JEE MCQ

Let us assume a force dF is applied at P in positive x - direction. This will stretch each spring by dl inducing a spring force dFs in each spring.

Let the static deformation of the system is dx (along the x-direction). The particle is in equilibrium. So,

Using Pythagoras theorem, 

Here y is constant. 

For the right (half) oscillation,

For the left (half) oscillation,

 absence of wall W2, the time period of block would have been  But due to the presence of W₂, it alters. But for the left part of oscillation (from the mean position shown), the period will be 

For the right side part, d = A sinω t [from x = A sinω t ] 

This is the time taken by the block to reach W₂ from mean position. Collision with W₂ is perfectly elastic (given). 

Time taken for right side part of oscillation will be 

∴ Total time period is 

∴ T ∝ √L , as time period decreases when the length of pendulum decreases, the time period of shorter pendulum (T_s) is smaller than that of longer pendulum (T_l). That means shorter pendulum performs more oscillations in a given time.

  It is given that after n oscillations of shorter pendulum, both are again in phase. So, by this time longer pendulum must have made (n – 1) oscillations. 

So, the two pendulum shall be in the same phase for the first time when the shorter pendulum has 
completed 5/4  oscillation  

When 500 g is removed, m = (100 + 300)g = 0.4 kg 

When 300 g is also removed, 

Total mechanical energy is 160
E_T = J
∴ U _max = 160 J

At extreme position KE is zero. Work done by spring force from extreme position to mean position is 

For x<0
F = −dU/dx = −kx
ma = −kx
or a = −kx/m
−ω^2₁x = −kx/m
ω₁ = (√k/m)
T₁ = 2π(√m/k)
For x>0 U = mgx
F = −dv/dx = −mg
But E = 1/2mv²₀
v₀ = (√2E/m)
It is speed at lowest point
T₂ = 2v₀/g
= 2/g(√2E/m)
T = T₁/2 + T₂
= π(√m/k) + 2/g(√2E/m)

Given:
 Mass of the particle, m = 0.1 kg
 Amplitude of SHM, A = 0.1 m
 Kinetic energy at mean position, K.E. = 8×10⁻³J
 Initial phase of oscillation, ϕ = 45^∘

The kinetic energy at the mean position is given by the formula:
KE = 1/2 mv²
At the mean position, the velocity v is maximum and is given by:
v_max = Aω
Substituting this into the kinetic energy formula gives:
K.E. = 1/2 m (Aω)²

Substituting the known values into the kinetic energy equation:
8×10⁻³ = 1/2 × 0.1 × (0.1ω)²
This simplifies to:
8 × 10⁻³ = 0.005 × (0.01ω²)
8 × 10⁻³ = 5 × 10⁻⁵ ω²
Now, solving for ω²:
ω² = 8 × 10⁻³ / 5 × 10⁻⁵ = 160
Thus,
ω = √160 = 4 radians/second

The general equation of motion for SHM is given by:
x(t) = A sin(ωt+ϕ)
Substituting the values of A, ω, and ϕ:
Convert ϕ from degrees to radians:
ϕ = 45^∘ = π/4 radians
Thus, the equation becomes:
x(t) = 0.1 sin(4t+π/4)

In 9s, the distance travelled = 4A × 3 = 12A 

Imagine, now it is at

Let particle (1) is moving towards right and particle (2) is moving towards left art this instant, 1 = 0 

In equilibrium position, mg = k x₀

When the bullet gets embedded, the new equilibrium position changes.  Let it is a distance x from initial equilibrium position. 

Applying conservation of linear momentum, 

So, at t = 0, mass 3m/2   is at a distance mg/2k  from it mean position and moving up with velocity u/3