NEET Oscillation MCQs & PYQs

To find the time required for a particle in Simple Harmonic Motion (SHM) to travel from \( x = A \) to \( x = \frac{A}{2} \), we can use the following steps:

1. Angular Frequency (\( \omega \)):
\[
\omega = \frac{2\pi}{T}
\]

2. Displacement in SHM:
The position \( x(t) \) in SHM can be described by:
\[
x(t) = A \cos(\omega t)
\]

3. Finding Time for Positions:
- For \( x = A \):
\[
A \cos(\omega t_1) = A;  \cos(\omega t_1) = 1 ; t_1 = 0
\]

- For \( x = \frac{A}{2} \):
\[
\frac{A}{2} = A \cos(\omega t_2 ; \cos(\omega t_2) = \frac{1}{2} ; \omega t_2 = \frac{\pi}{3}
\]
\[
t_2 = \frac{\pi}{3\omega} = \frac{\pi T}{6}
\]

4. Time Interval:
The time taken to travel from \( x = A \) to \( x = \frac{A}{2} \) is:
\[
\Delta t = t_2 - t_1 = \frac{\pi T}{6} - 0 = \frac{T}{6}
\]

Thus, the time required to travel from \( x = A \) to \( x = \frac{A}{2} \) is:
\[{\frac{T}{6}}
\]

In Simple Harmonic Motion (SHM), the relationship between displacement, velocity, and amplitude can be described using the following equations.

1. The maximum speed \( V_0 \) is given by:
\[
V_0 = \omega A
\]
where \( \omega \) is the angular frequency.

2. The speed \( v \) at a displacement \( x \) in SHM is given by the formula:
\[
v = \sqrt{V_0^2 - \left(\frac{\omega x}{\omega A}\right)^2}
\]
Simplifying this using \( \omega = \frac{V_0}{A} \):
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{A} \cdot x\right)^2}
\]
Substituting \( x = \frac{A}{2} \):
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{A} \cdot \frac{A}{2}\right)^2}
\]
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{2}\right)^2}
\]
\[
v = \sqrt{V_0^2 - \frac{V_0^2}{4}} = \sqrt{\frac{3V_0^2}{4}} = \frac{\sqrt{3}}{2} V_0
\]

Thus, the speed of the particle at displacement \( \frac{A}{2} \) is:
\[{\frac{\sqrt{3}}{2} V_0}
\]

In simple harmonic motion (SHM), the maximum velocity \( v_{\text{max}} \) and maximum acceleration \( a_{\text{max}} \) are given by:

\[
v_{\text{max}} = \omega A \quad \text{and} \quad a_{\text{max}} = \omega^2 A
\]

where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude.

 Solution:

1. Given that \( v_{\text{max}} = a_{\text{max}} \), we have:
\[
\omega A = \omega^2 A
\]

2. Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[
\omega = \omega^2
\]

3. Solving for \( \omega \):
\[
\omega = 1 \, \text{rad/s}
\]

4. Time Period \( T \):
\[
T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi \approx 6.28 \, \text{s}
\]

Answer:
The time period of the particle is **6.28 seconds**.

For the SHM given by:

\[
y = A \cos(\omega t + \frac{\pi}{4})
\]

Solution:

1. Velocity: The velocity \( v \) is the derivative of \( y \) with respect to \( t \):
\[
v = \frac{dy}{dt} = -A \omega \sin(\omega t + \frac{\pi}{4})
\]

2. Maximum Speed: The speed will be maximum when \( \sin(\omega t + \frac{\pi}{4}) = \pm 1 \).

Therefore,
\[
\omega t + \frac{\pi}{4} = \frac{\pi}{2}
\]

3. Solving for \( t \):
\[
\omega t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
\]

\[
t = \frac{\pi}{4\omega}
\]

Answer:
The speed will be maximum at \( t = \frac{\pi}{4\omega} \).

In simple harmonic motion (SHM), the velocity \( v \) and displacement \( x \) of a particle are related by:

\[
v = \pm \sqrt{\omega^2 A^2 - \omega^2 x^2}
\]

where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude.

 Solution:

1. Rearranging this equation, we get:
\[
\frac{v^2}{\omega^2 A^2} + \frac{x^2}{A^2} = 1
\]

2. This is the equation of an ellipse in the \( v \)-\( x \) plane.

Answer:
The graph of velocity as a function of displacement for a particle in SHM is an ellipse.

Given the SHM equation:

\[
2 \frac{d^2x}{dt^2} + 32x = 0
\]

We can rewrite it as:

\[
\frac{d^2x}{dt^2} + 16x = 0
\]

This equation is of the form:

\[
\frac{d^2x}{dt^2} + \omega^2 x = 0
\]

where \( \omega^2 = 16 \).

Solution:

1. Angular Frequency \( \omega \):
\[
\omega = \sqrt{16} = 4 \, \text{rad/s}
\]

2. Time Period \( T \):
\[
T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s}
\]

 Answer:
The time period of the oscillation is \( \frac{\pi}{2} \) seconds.

Solution:

1. Maximum Acceleration in SHM is given by:
\[
a_{\text{max}} = \omega^2 A
\]

2. Ratio of Maximum Accelerations:
\[
\frac{a_{\text{max}_2}}{a_{\text{max}_1}} = \frac{\omega_2^2 A}{\omega_1^2 A} = \frac{\omega_2^2}{\omega_1^2} = \frac{(100)^2}{(10)^2} = \frac{10000}{100} = 100
\]

Answer:
The ratio of their maximum accelerations is \( 1 : 100 \) or \( 1 : 10^2 \).

Given:
- Amplitude, \( A = 0.01 \, \text{m} \)
- Frequency, \( f = 60 \, \text{Hz} \)

Solution:

1. Angular Frequency \( \omega \):
\[
\omega = 2 \pi f = 2 \pi \times 60 = 120 \pi \, \text{rad/s}
\]

2. Maximum Acceleration \( a_{\text{max}} \):
Maximum acceleration in SHM is given by:
\[
a_{\text{max}} = \omega^2 A
\]

Substitute values of \( \omega \) and \( A \):
\[
a_{\text{max}} = (120 \pi)^2 \times 0.01 = 14400 \pi^2 \times 0.01 = 144 \pi^2 \, \text{m/s}^2
\]

Answer:
The maximum acceleration of the particle is \( 144 \pi^2 \, \text{m/s}^2 \).

Given the SHM equation:

\[
x = 4 \cos(88t + \frac{\pi}{4})
\]

Solution:

1. Frequency: The general form of SHM is \( x = A \cos(\omega t + \phi) \), where \( \omega = 2 \pi f \) (angular frequency).

Here, \( \omega = 88 \).

\[
f = \frac{\omega}{2 \pi} = \frac{88}{2 \pi} = 14 \, \text{Hz}
\]

2. Initial Displacement: The initial displacement \( x_0 \) is the value of \( x \) at \( t = 0 \).

Substitute \( t = 0 \) into the equation:
\[
x_0 = 4 \cos\left(\frac{\pi}{4}\right) = 4 \cdot \frac{\sqrt{2}}{2} = 2 \sqrt{2} \, \text{m}
\]

Answer:
The frequency is  14 Hz, and the initial displacement is \( 2 \sqrt{2} \, \text{m} \).

As displacement in one time interval is zero. Average velocity is zero.