Acceleration and Force of Simple Harmonic Motion Questions in English

(A) The displacement of the particle is given by $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = -x_0 \omega \sin \left(\omega t-\frac{\pi}{4}\right)$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}\right)$.
Using the trigonometric identity $\cos(\theta + \pi) = -\cos(\theta)$,we can rewrite the acceleration as:
$a = x_0 \omega^2 \cos \left(\omega t - \frac{\pi}{4} + \pi\right) = x_0 \omega^2 \cos \left(\omega t + \frac{3\pi}{4}\right)$.
To match the form $a = A \cos(\omega t - \delta)$,we write the phase as $\omega t - (-\frac{3\pi}{4})$.
Thus,$A = x_0 \omega^2$ and $\delta = -\frac{3\pi}{4}$.

(A) The displacement equation for simple harmonic motion is given as $x = 6 \cos \left(2 \pi t + \frac{\pi}{3}\right)$.
The velocity $v$ is the first derivative of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt} \left[6 \cos \left(2 \pi t + \frac{\pi}{3}\right)\right] = -6 \sin \left(2 \pi t + \frac{\pi}{3}\right) \times 2 \pi = -12 \pi \sin \left(2 \pi t + \frac{\pi}{3}\right)$.
The acceleration $a$ is the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt} \left[-12 \pi \sin \left(2 \pi t + \frac{\pi}{3}\right)\right] = -12 \pi \cos \left(2 \pi t + \frac{\pi}{3}\right) \times 2 \pi = -24 \pi^2 \cos \left(2 \pi t + \frac{\pi}{3}\right)$.
At $t = 1 \ s$,the acceleration is:
$a = -24 \pi^2 \cos \left(2 \pi(1) + \frac{\pi}{3}\right) = -24 \pi^2 \cos \left(2 \pi + \frac{\pi}{3}\right)$.
Since $\cos(2 \pi + \theta) = \cos \theta$,we have $\cos \left(2 \pi + \frac{\pi}{3}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$.
Therefore,$a = -24 \pi^2 \times \frac{1}{2} = -12 \pi^2 \ m/s^2$.
The magnitude of acceleration is $|a| = 12 \pi^2 \ m/s^2$.

(B) The displacement of the body is given by $x = 3 \cos \left( 2 \pi t + \frac{\pi}{4} \right)$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = -3 \sin \left( 2 \pi t + \frac{\pi}{4} \right) \cdot (2 \pi) = -6 \pi \sin \left( 2 \pi t + \frac{\pi}{4} \right)$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -6 \pi \cos \left( 2 \pi t + \frac{\pi}{4} \right) \cdot (2 \pi) = -12 \pi^2 \cos \left( 2 \pi t + \frac{\pi}{4} \right)$.
At $t = 2 \text{ s}$,the acceleration is $a = -12 \pi^2 \cos \left( 2 \pi (2) + \frac{\pi}{4} \right) = -12 \pi^2 \cos \left( 4 \pi + \frac{\pi}{4} \right)$.
Since $\cos(4 \pi + \theta) = \cos \theta$,we have $a = -12 \pi^2 \cos \left( \frac{\pi}{4} \right) = -12 \pi^2 \cdot \frac{1}{\sqrt{2}} = -6 \sqrt{2} \pi^2 \text{ m/s}^2$.

(B) The maximum acceleration of a simple harmonic oscillator is given by $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
The maximum velocity is given by $v_{max} = \omega A$.
According to the problem,the magnitude of maximum acceleration is $\pi$ times the maximum velocity:
$a_{max} = \pi \cdot v_{max}$
Substituting the expressions:
$\omega^2 A = \pi \cdot \omega A$
Dividing both sides by $\omega A$ (assuming $\omega, A \neq 0$):
$\omega = \pi$
We know that the angular frequency $\omega$ is related to the time period $T$ by the formula $\omega = \frac{2\pi}{T}$.
Substituting $\omega = \pi$:
$\pi = \frac{2\pi}{T}$
$T = 2 \text{ s}$.

(D) The given displacement equation for $SHM$ is $y = 2 \sin \left(\frac{\pi t}{2} + \phi\right) \text{ cm}$.
Comparing this with the standard $SHM$ equation $y = A \sin(\omega t + \phi)$,we get the amplitude $A = 2 \text{ cm}$ and angular frequency $\omega = \frac{\pi}{2} \text{ rad/s}$.
The formula for maximum acceleration in $SHM$ is $a_{\max} = \omega^2 A$.
Substituting the values,we get $a_{\max} = \left(\frac{\pi}{2}\right)^2 \times 2$.
$a_{\max} = \frac{\pi^2}{4} \times 2 = \frac{\pi^2}{2} \text{ cm/s}^2$.

(A) In simple harmonic motion,the acceleration $f$ is given by $f = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.
Taking the magnitude,we have $|f| = \omega^2 |x|$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
Therefore,the product $|fT|$ is:
$|fT| = |f| \cdot T = (\omega^2 |x|) \cdot \left(\frac{2\pi}{\omega}\right) = 2\pi\omega |x|$.
Since $2\pi\omega$ is a constant,the relationship $|fT| = (2\pi\omega) |x|$ represents a linear equation of the form $y = mx$,where $y = |fT|$ and $x$ is the displacement.
Thus,the graph of $|fT|$ versus $x$ is a straight line passing through the origin.

(C) In Simple Harmonic Motion $(SHM)$,the acceleration $a$ of a vibrating particle is given by the relation $a = \omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.
Given: Acceleration $a = 16 \ cm/s^2$ and displacement $x = 4 \ cm$.
Substituting these values into the formula:
$16 = \omega^2 \times 4$
$\omega^2 = \frac{16}{4} = 4$
$\omega = 2 \ rad/s$
The time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi \ s$
Using $\pi \approx 3.142$,we get $T = 3.142 \ s$.