Test: Periodic Oscillatory & Simple Harmonic Motion (September 23) - NEET MCQ

For the right (half) oscillation,

For the left (half) 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 

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)

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