31 Year NEET Previous Year Questions: Oscillations - 1 - NEET MCQ

At any point of time, time period is given by
T = 2π √(m / k)
Here m is decreasing, so time period T will be decreasing
Since ω = 2π / T
Hence as mass leaks, ω will increase
Now, at any instant
mg = kx₀
So, equilibrium length x₀ = mg / k, where m is decreasing

So, equilibrium length will decrease.
So, amplitude also go on decreasing.

Maximum speed in simple harmonic motion is given by:
v_max = A × ω
Since the masses are identical, angular frequency ω is given by:
ω = √(k / m)
Let A_P and A_Q be amplitudes and ω₁ and ω₂ be angular frequencies of P and Q respectively.

Given: v_max(P) = v_max(Q)
⇒ A_P × ω₁ = A_Q × ω₂
⇒ A_Q / A_P = ω₁ / ω₂
⇒ A_Q / A_P = √(k₁) / √(k₂)
⇒ A_Q / A_P = √(k₁ / k₂)
Therefore, the correct answer is Option B: √(k₁ / k₂)

The period of oscillation, T, of a simple pendulum is determined by the formula:

where:

• L is the length of the pendulum
• g is the acceleration due to gravity

The mass of the bob does not factor into the equation for the period.
Let's first denote the original length of the pendulum as L and the original period of oscillation as T₁. Hence,

When the length of the pendulum is halved, the new length L' would be L / 2. Thus, the new period T₂ can be calculated as:

We are given that the new period T₂2 is . Therefore, we can set up the equation:

To find the value of x, we solve for x:

Multiplying both sides by 2:

Simplify to:
x =√2
Hence, the correct answer is:
Option B: √2

In the equation for simple harmonic motion (SHM), , the general form  can be used to identify the parameters of SHM, where:

• A is the amplitude.
• ω (omega) is the angular frequency.
• φ (phi) is the phase constant.
• t is the time.

Comparing the given equation with the standard form:

• The amplitude A is 5 m, as that is the coefficient of sine in the equation.
• The angular frequency ω isπ rad/s.

The angular frequency ω is related to the time period T of the motion through the formula:

Given that ω=π, we can substitute and solve for T:

Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is:
Option B: 5 m, 2 s

Position of particle as function of time

(a) Amplitude of oscillation is continuously decreasing. It means bob of pendulum oscillate with air friction.

(b) F = -kx (restoring force of a spring) 

(c) Amplitude of oscillation is remains same. It means bob of pendulum is oscillating under negligible air resistance.

(d)  

A periodic function repeats its value after a fixed time interval.
(a) sin(ωt + π/4)
Sine function is periodic with period = 2π/ω.
So, it represents periodic motion.
(b) e^-ωt
This is an exponential decay function.
It does not repeat its values with time.
Hence, it represents non-periodic motion.
(c) sin ωt
Sine function is periodic with period = 2π/ω.
So, it represents periodic motion.
(d) sin ωt + cos ωt
Sum of two periodic functions with the same angular frequency ω is also periodic.
So, it represents periodic motion.
Correct answer: (b) e^-ωt

Displacement equation of SHM of frequency 'n' x = Asin(ωt) = Asin(2πnt) Now,

So frequency of potential energy = 2n

F = kx
10 = k(5 × 10^–2)

= 2 x 10²
= 200 N/m
Now,

At t = 0, y displacement is maximum, so equation will be cosine function.

T = 4s

y = a cosωt

y = A₀+ Asinωt + Bsinωt
Equate SHM
y' = y – A₀ = Asinωt + Bcosωt
Resultant amplitude

y = A₀+ Asinωt + Bsinωt
Equate SHM
y' = y – A₀ = Asinωt + Bcosωt
Resultant amplitude

In one complete vibration, displacement is zero. So, average velocity in one complete vibration

Let us assume, the length of spring be l.
When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths  with force constant,

When connected in series,

When connected in parallel,

The time period of oscillation is given by

On dividing :

As, we know, in SHM

where c² = a² + b²  [since  a² + b² = c² cos (φ) + c² sin (φ) = c² ]

The superimposed motion is simple harmonic with amplitude 

This is the equation ofellipse. Hence the graph is an ellipse. P versus x graph is similar to V versus x graph.

on comparing with the standard wave equation

y = a sin (ωt – kx) Wave velocity

The velocity of particle

Equation of SHM is given by x = A sin (ωt + δ) (ωt + δ) is called phase.

For second particle,