A particle executes linear simple harmonic motion with an amplitude of 2 cm. when the particle is at 1 cm from the mean position, the magnitude of the velocity and the acceleration are equal. Then its item period (in seconds) is
Understanding Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is directly proportional to the displacement from the mean position and acts towards the mean position. We are given a particle executing linear SHM.
Analyzing the Given SHM Problem
We are provided with the following information for the particle's motion:
- Amplitude of the SHM, \(A = 2\) cm.
- Position of the particle from the mean position, \(x = 1\) cm.
- At this specific position (\(x = 1\) cm), the magnitude of the particle's velocity is equal to the magnitude of its acceleration. That is, \(|v| = |a|\).
We need to find the time period (\(T\)) of the oscillation in seconds.
Formulas for Velocity and Acceleration in SHM
For a particle executing SHM, the velocity (\(v\)) and acceleration (\(a\)) at a displacement \(x\) from the mean position are given by:
- Velocity: \(v = \pm \omega \sqrt{A^2 - x^2}\), where \(\omega\) is the angular frequency.
- Acceleration: \(a = -\omega^2 x\), where \(\omega\) is the angular frequency.
Setting up the Equation from the Given Condition
The problem states that the magnitude of velocity equals the magnitude of acceleration when \(x = 1\) cm.
\(|v| = |a|\)
Using the formulas:
\(|\pm \omega \sqrt{A^2 - x^2}| = |-\omega^2 x|\)
Taking magnitudes, and knowing \(\omega\) is a positive value for oscillations and \(x=1\) cm is a positive displacement:
\(\omega \sqrt{A^2 - x^2} = \omega^2 x\)
Solving for Angular Frequency (\(\omega\))
Now, we substitute the given values \(A = 2\) cm and \(x = 1\) cm into the equation:
\(\omega \sqrt{(2)^2 - (1)^2} = \omega^2 (1)\)
\(\omega \sqrt{4 - 1} = \omega^2\)
\(\omega \sqrt{3} = \omega^2\)
Since the particle is oscillating, \(\omega\) cannot be zero. We can divide both sides by \(\omega\):
\(\sqrt{3} = \omega\)
So, the angular frequency is \(\omega = \sqrt{3}\) rad/s.
Calculating the Time Period (T) of SHM
The relationship between angular frequency (\(\omega\)) and time period (\(T\)) is given by:
\(T = \frac{2\pi}{\omega}\)
Substituting the value of \(\omega = \sqrt{3}\):
\(T = \frac{2\pi}{\sqrt{3}}\) seconds
Conclusion
The time period of the particle executing simple harmonic motion is \(\frac{2\pi}{\sqrt{3}}\) seconds.
| Given Information | Value |
|---|---|
| Amplitude (A) | 2 cm |
| Position (x) | 1 cm |
| Condition | \(|v| = |a|\) at \(x=1\) cm |
| Derived Value | Value |
|---|---|
| Angular Frequency (\(\omega\)) | \(\sqrt{3}\) rad/s |
| Time Period (T) | \(\frac{2\pi}{\sqrt{3}}\) s |
SHM Problem Revision Table
Let's quickly review the key formulas used in this SHM problem.
| Concept | Formula |
|---|---|
| Velocity (v) at position x | \(v = \pm \omega \sqrt{A^2 - x^2}\) |
| Acceleration (a) at position x | \(a = -\omega^2 x\) |
| Relationship between T and \(\omega\) | \(T = \frac{2\pi}{\omega}\) |
Additional Information on Simple Harmonic Motion
Simple Harmonic Motion is a fundamental concept in physics, describing oscillations. Here's a bit more detail:
- Mean Position: This is the equilibrium position where the net force on the particle is zero. Displacement (x) is measured from this position.
- Amplitude (A): The maximum displacement of the particle from its mean position.
- Angular Frequency (\(\omega\)): Represents how fast the oscillation is occurring, related to the properties of the system (like mass and spring constant for a mass-spring system). Units are typically radians per second.
- Time Period (T): The time taken for one complete oscillation (or cycle).
- Frequency (f): The number of oscillations per unit time. \(f = 1/T = \omega / (2\pi)\).
- Relationship between v, a, x, A, \(\omega\): The formulas used in the solution (\(v = \pm \omega \sqrt{A^2 - x^2}\) and \(a = -\omega^2 x\)) are derived from the basic differential equation of SHM, \(\frac{d^2x}{dt^2} = -\omega^2 x\).
Understanding these fundamental concepts is crucial for solving problems related to SHM, including those involving velocity, acceleration, amplitude, and time period.
Similar Questions
A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be:
(in terms of period T of the first pendulum)
Important Questions from Simple Harmonic Motion
- The displacement of a particle is given by $y(t) = K + P \sin^2(\omega t) + Q \sin(\omega t) \cos(\omega t)$. If this represents a simple harmonic motion, the amplitude of its oscillation is:
- Which one of the following equations of motion represents simple harmonic motion?
Assume $A$, $B$, $C$, $D$, $m$, $k$, and $\omega$ are all positive constants. In simple harmonic motion, the particle velocity lags behind the displacement by a phase angle of __________.
- A particle undergoes simple harmonic motion. Determine the phase difference between its instantaneous velocity and instantaneous acceleration.
- A particle executes simple harmonic motion along a straight line. When its displacement from the mean position is $x$, its speed is $v$. If the displacement becomes $2x$, its speed reduces to $v/2$. What is the amplitude of the oscillation?
