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Q13.10 (b):
Let us take the position of mass when the spring is unstreched as (x = 0), and the direction from left to right as the positive direction of (x)-axis. Give (x) as a function of time (t) for the oscillating mass if at the moment we start the stopwatch ((t = 0)), the mass is at the maximum stretched position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Solution :

Step 1: Establishing the General Equation of SHM

The general displacement equation for a particle performing Simple Harmonic Motion (SHM) is given by:

$x(t) = A \cos(\omega t + \phi)$

Where:

  • $A$ is the amplitude of oscillation.
  • $\omega$ is the angular frequency, defined as $\omega = \sqrt{\frac{k}{m}}$.
  • $\phi$ is the initial phase constant (phase angle at $t = 0$).
[Per the general solution of the second-order linear differential equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$]

Step 2: Applying Initial Conditions

The problem states that at $t = 0$, the mass is at its maximum stretched position. In the defined coordinate system where right is positive and $x=0$ is the equilibrium position, the maximum stretch occurs at $x = +A$.

Substitute the initial condition into the general equation:

$x(0) = A \cos(\omega(0) + \phi) = A$

$A \cos(\phi) = A$

$\cos(\phi) = 1$

$\phi = 0$ radians.

Step 3: Deriving the Specific Function of Time

Substituting $\phi = 0$ back into the general equation:

$x(t) = A \cos(\omega t + 0)$

$x(t) = A \cos(\omega t)$

Step 4: Comparison Analysis

If we compare this function to a situation where the mass starts from the mean position ($x=0$ at $t=0$), that function would be $x(t) = A \sin(\omega t)$ or $x(t) = A \cos(\omega t - \frac{\pi}{2})$.

Parameter Comparison Result
Frequency ($\omega$) Identical; both depend solely on spring constant ($k$) and mass ($m$).
Amplitude ($A$) Identical; determined by the initial energy/displacement provided to the system.
Initial Phase ($\phi$) Different; the phase differs by $\frac{\pi}{2}$ radians ($90^\circ$).

Visual Representation of the Displacement

t x(t) +A -A Graph of x(t) = A cos(ωt)

Final Solution: The function is $x(t) = A \cos(\omega t)$. These functions differ from each other only in the initial phase ($\phi$), while the frequency and amplitude remain identical as they are intrinsic properties of the physical oscillator.


More Questions from Class 11 Physics Oscillations Exercises


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