Test: Simple Harmonic Motion - NEET MCQ

Equation of SHM particle: 
Y=a sinωt
V=aω sinωt
V_max = aω
So the velocity is maximum at mean position

Periodic time is the time taken for one complete revolution of the particle.

∴ Periodic time (t_p) = 2 π/ω seconds.

Frequency is defined as time taken to perform one oscillation by the object.

Hence, 1Hz corresponds to 1 vibration per sec.

• Time Period, T = 2π √(l/g')where,
  l = Length of seconds pendulum 
  g’ = Apparent Gravity
• For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
• Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.

The given potential field is V = (50x² + 100) J/kg.

For a particle of mass m undergoing simple harmonic motion, the potential energy can be expressed as V = (1/2) k x², where k is the force constant. Comparing this with the given potential, we identify (1/2) m ω² = 50. Here, m = 10 gm = 0.01 kg.

Solving for ω, the angular frequency, we get ω = sqrt(100/0.01) = 100 rad/s.

The frequency f in cycles per second is given by f = ω/2π = 100/2π = 50/π.

We know that in a simple harmonic motion the maximum velocity,
V_max = A⍵
Here A = 50 mm

And ⍵ = 2π / T
= 2π / 2
= π

Hence  V_max = 50 x 10^-3.π
= 0.15 m/s

When the displacement of a SHM is:
y=a sin wt+ b cos wt

• Amplitude of the SHM will be:
  A=√a²+b²

Here, a = 3, b = 4
Amplitude, A= √(3²+4²) = 5 cm

Hence option B is correct.

After one oscillation of a bigger pendulum i.e. 5T/4, ¼ of the total phase is travelled by the smaller pendulum while the bigger is still at initial position.

Thus, the phase difference between two is: ¼ (2π) - 0 = π/2

   ∴ We get, ω = √3 s^-1
   T = 2π / √3