Acceleration and Force of Simple Harmonic Motion Questions in English

(B) Given: Mass $m = 0.3 \, kg$,force constant $k = 15 \, N/m$,and displacement $x = 20 \, cm = 0.2 \, m$.
According to Hooke's Law,the magnitude of the force is $F = kx$.
Substituting the values: $F = 15 \times 0.2 = 3 \, N$.
Using Newton's Second Law,$F = ma$,the acceleration $a$ is given by $a = \frac{F}{m}$.
Substituting the values: $a = \frac{3}{0.3} = 10 \, m/s^2$.
Therefore,the initial acceleration is $10 \, m/s^2$.

(C) The acceleration $a$ of a particle in simple harmonic motion is given by $a = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the displacement.
Given $a = 0.5 \, m/s^2$ and $y = 0.02 \, m$.
$\omega^2 = \frac{a}{y} = \frac{0.5}{0.02} = 25$.
$\omega = \sqrt{25} = 5 \, rad/s$.
The maximum velocity $v_{\max}$ is given by $v_{\max} = A\omega$,where $A$ is the amplitude.
Given $A = 0.1 \, m$.
$v_{\max} = 0.1 \times 5 = 0.5 \, m/s$.

(C) The acceleration $a$ of a particle executing $S.H.M.$ is given by the formula $a = -{\omega ^2}x$,where $\omega$ is the angular frequency and $x$ is the displacement from the equilibrium position.
At the equilibrium position,$x = 0$,so the acceleration is zero.
At the extreme positions,the displacement $x$ is equal to the amplitude $A$ (i.e.,$x = \pm A$).
Therefore,the magnitude of acceleration is $|a| = {\omega ^2}A$,which is the maximum value.
Thus,the acceleration is maximum at the extreme position.

(D) The displacement of the particle is given by $y = a \sin \omega t$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dy}{dt} = a \omega \cos \omega t$.
The acceleration $A$ is the derivative of velocity with respect to time: $A = \frac{dv}{dt} = -a \omega^2 \sin \omega t$.
Given $t = \frac{T}{4}$ and knowing $\omega = \frac{2\pi}{T}$,we have $\omega t = \frac{2\pi}{T} \times \frac{T}{4} = \frac{\pi}{2}$.
Substituting this into the acceleration equation: $A = -a \omega^2 \sin(\frac{\pi}{2}) = -a \omega^2 (1) = -a \omega^2$.
Thus,the correct option is $D$.

(A) The maximum acceleration $a_{max}$ of a particle executing $S.H.M.$ is given by the formula $a_{max} = A\omega^2$.
Here,$A = 0.01 \, m$ is the amplitude and $n = 60 \, Hz$ is the frequency.
Since $\omega = 2\pi n$,we have $\omega^2 = 4\pi^2 n^2$.
Substituting the values: $a_{max} = 0.01 \times 4\pi^2 \times (60)^2$.
$a_{max} = 0.01 \times 4\pi^2 \times 3600$.
$a_{max} = 144\pi^2 \, m/s^2$.

(A) The maximum force in $S.H.M.$ is given by $F_{\max} = m \cdot a_{\max}$,where $a_{\max} = A \omega^2$.
Given: mass $m = 0.10 \, kg$,amplitude $A = 1.0 \, m$,and period $T = 0.20 \, s$.
The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.20} = 10\pi \, rad/s$.
The maximum acceleration is $a_{\max} = A \omega^2 = 1.0 \times (10\pi)^2 = 100\pi^2 \, m/s^2$.
The maximum force is $F_{\max} = m \cdot a_{\max} = 0.10 \times 100\pi^2 = 10\pi^2 \, N$.
Using $\pi^2 \approx 9.8596$,we get $F_{\max} = 10 \times 9.8596 = 98.596 \, N$.

(D) In simple harmonic motion $(SHM)$,the acceleration $a$ is given by $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the equilibrium position.
At the equilibrium position,$x = 0$,so the acceleration $a = 0$,which is the minimum value.
At the extreme positions,$x = \pm A$ (where $A$ is the amplitude),so the acceleration $a = \mp \omega^2 A$,which is the maximum value.
Therefore,the statement that the acceleration is maximum at the equilibrium position is incorrect.
The correct option is $D$.

(D) The mass of the particle is $m = 10 \text{ g} = 0.01 \text{ kg}$.
The amplitude is $A = 0.5 \text{ m}$.
The time period is $T = \pi / 5 \text{ s}$.
The angular frequency is $\omega = 2\pi / T = 2\pi / (\pi / 5) = 10 \text{ rad/s}$.
The maximum force acting on a particle in simple harmonic motion is given by $F_{\text{max}} = m \omega^2 A$.
Substituting the values: $F_{\text{max}} = 0.01 \times (10)^2 \times 0.5$.
$F_{\text{max}} = 0.01 \times 100 \times 0.5 = 1 \times 0.5 = 0.5 \text{ N}$.

(D) The given equation for displacement is $y = \sin \left( \frac{\pi t}{4} + \frac{\pi}{6} \right)$.
Comparing this with the standard $SHM$ equation $y = A \sin(\omega t + \phi)$,we get amplitude $A = 1 \ cm$ and angular frequency $\omega = \frac{\pi}{4} \ rad/s$.
The acceleration $a$ of a particle in $SHM$ is given by $a = -\omega^2 y$.
The maximum acceleration is given by $a_{\max} = \omega^2 A$.
Substituting the values: $a_{\max} = \left( \frac{\pi}{4} \right)^2 \times 1 = \frac{\pi^2}{16}$.
Using $\pi^2 \approx 9.869$,we get $a_{\max} = \frac{9.869}{16} \approx 0.6168 \ cm/s^2$.
Rounding to two decimal places,we get $a_{\max} \approx 0.62 \ cm/s^2$.

(A) For a particle executing Simple Harmonic Motion $(S.H.M.)$,the restoring force is directly proportional to the negative of the displacement from the mean position.
Mathematically,this is expressed as $F = -kx$,where $k$ is the force constant.
Given that $A$ and $K$ are positive constants,the force equation for this specific motion is $F = -AKx$.
Therefore,the correct option is $A$.

(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = A\omega^2$,where $A$ is the amplitude and $\omega$ is the angular velocity.
Given:
Angular velocity $\omega = 3.5\, rad/s$
Maximum acceleration $a_{\max} = 7.5\, m/s^2$
Rearranging the formula to solve for amplitude $A$:
$A = \frac{a_{\max}}{\omega^2}$
Substituting the given values:
$A = \frac{7.5}{(3.5)^2} = \frac{7.5}{12.25} \approx 0.61\, m$
Therefore,the amplitude of oscillation is $0.61\, m$.

(A) The displacement equation is given by $x = A \cos(\omega t + \phi)$,where $A = 10\, cm = 0.10\, m$ and $\omega = 10\, rad/s$.
The acceleration $a$ of a particle in simple harmonic motion is given by $a = -\omega^2 x$.
The maximum acceleration $a_{\max}$ is given by the formula $a_{\max} = \omega^2 A$.
Substituting the given values: $a_{\max} = (10\, rad/s)^2 \times (0.10\, m) = 100 \times 0.10 = 10\, m/s^2$.
Therefore,the maximum acceleration experienced by the block is $10\, m/s^2$.

(A) The acceleration of a particle in $S.H.M.$ is given by $a = -{\omega ^2}x$,where $x$ is the displacement from the equilibrium position.
Maximum acceleration occurs when the displacement $x$ is maximum,which is equal to the amplitude $A$.
Therefore,$|a_{\max}| = {\omega ^2}A$.
This occurs at the extreme positions,i.e.,at the amplitude.

(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = \omega^2 A$,where $\omega = 2\pi f$.
Given: Amplitude $(A)$ = $0.02 \ m$,Frequency $(f)$ = $50 \ Hz$.
First,calculate the angular frequency: $\omega = 2 \times \pi \times 50 = 100\pi \ rad/s$.
Now,calculate the maximum acceleration: $a_{\max} = (100\pi)^2 \times 0.02$.
$a_{\max} = 10000 \pi^2 \times 0.02 = 200 \pi^2 \ m/s^2$.

(D) The displacement of a particle executing $SHM$ is given by $x = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
The acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x$.
At the mean position,the displacement $x = 0$.
Substituting $x = 0$ into the acceleration formula,we get $a = -\omega^2(0) = 0$.
Therefore,the acceleration of a particle executing $SHM$ at its mean position is zero.

(D) In simple harmonic motion $(S.H.M.)$,the displacement is given by $y = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi)$.
The acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 y$.
At the mean position $(y = 0)$,the speed $v$ is maximum $(v_{max} = A\omega)$ and the acceleration $a$ is zero $(a = -\omega^2(0) = 0)$.
At the extreme positions $(y = \pm A)$,the speed $v$ is zero and the acceleration $a$ is maximum $(a_{max} = \pm \omega^2 A)$.
Therefore,when $v$ is maximum,$a$ is zero.

(B) Comparing the given equation $y = 2\sin \left( {\frac{{\pi t}}{2} + \phi } \right)$ 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 maximum acceleration $a_{max}$ is given by the formula $a_{max} = \omega^2 A$.
Substituting the values: $a_{max} = \left( \frac{\pi }{2} \right)^2 \times 2 = \frac{\pi^2}{4} \times 2 = \frac{\pi^2}{2} \text{ cm/s}^2$.

(C) The acceleration $a$ of a particle executing simple harmonic motion is given by the equation $a = -{\omega}^2 x$,where ${\omega}$ is the angular frequency and $x$ is the displacement.
Taking the magnitude of the ratio of acceleration to displacement,we get:
$|a/x| = |-{\omega}^2 x / x| = {\omega}^2$.
Therefore,the ratio of the acceleration to the displacement is equal to the square of the angular frequency.

(D) For a particle performing $S.H.M.$,the acceleration $a$ is given by $a = -\omega^2 x$,where $x$ is the displacement from the mean position and $\omega$ is the angular frequency.
Given: Displacement $x = 3\;cm$ and acceleration $a = 12\;cm/s^2$.
Taking the magnitude: $12 = \omega^2 \times 3$.
$\omega^2 = \frac{12}{3} = 4$.
$\omega = 2\;rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi\;s$.
Using $\pi \approx 3.14$,we get $T = 3.14\;s$.

(A) For a simple harmonic oscillator,the magnitude of acceleration $a$ is given by the relation $a = \omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.
Given: $x = 0.02\, m$ and $a = 2.0\, m\, s^{-2}$.
Rearranging the formula for angular frequency: $\omega = \sqrt{\frac{a}{x}}$.
Substituting the values: $\omega = \sqrt{\frac{2.0}{0.02}} = \sqrt{\frac{200}{2}} = \sqrt{100}$.
Therefore,$\omega = 10\, rad\, s^{-1}$.

(B) Given the displacement equation: $x = 12 \sin \omega t - 16 \sin^3 \omega t$.
Using the trigonometric identity $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$,we can rewrite the expression as:
$x = 4(3 \sin \omega t - 4 \sin^3 \omega t) = 4 \sin(3 \omega t)$.
This represents a $S.H.M.$ with amplitude $A = 4 \ cm$ and angular frequency $\omega' = 3 \omega$.
The acceleration $a$ in $S.H.M.$ is given by $a = -\omega'^2 x$.
The maximum acceleration is $a_{\max} = \omega'^2 A$.
Substituting the values: $a_{\max} = (3 \omega)^2 \times 4 = 9 \omega^2 \times 4 = 36 \omega^2 \ cm/s^2$.

(A) For a particle executing $S.H.M.$,the acceleration $a$ is given by the equation $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.
This equation is of the form $y = mx$,which represents a straight line passing through the origin.
Here,the slope of the line is $m = -\omega^2$,which is negative.
Therefore,the graph of acceleration versus displacement is a straight line with a negative slope.

(D) The acceleration of a particle in $S.H.M.$ is given by the relation $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.
For the particle to be at $-x_{max}$,we substitute $x = -x_{max}$ into the equation:
$a = -\omega^2 (-x_{max}) = +\omega^2 x_{max}$.
This shows that at $x = -x_{max}$,the acceleration $a$ must be at its maximum positive value.
Looking at the provided graph,point $1$ corresponds to the maximum positive value of acceleration.
Therefore,point $1$ corresponds to the particle being at $-x_{max}$.

(D) For a particle executing $S.H.M.$,the displacement $y$ is given by $y = A \sin(\omega t)$.
Acceleration $a = \frac{d^2y}{dt^2} = -A\omega^2 \sin(\omega t) = -\omega^2 y$.
According to Newton's second law,the restoring force $F$ is given by $F = ma = -m\omega^2 y$.
Substituting the expression for $y$,we get $F = -m\omega^2 A \sin(\omega t)$.
This shows that the force $F$ is also a sinusoidal function of time but with a phase difference of $\pi$ radians relative to the displacement $y$.
Therefore,if the displacement graph is a positive sine wave starting from the origin,the force-time graph must be an inverted (negative) sine wave starting from the origin,which corresponds to option $D$.

(C) In Simple Harmonic Motion $(S.H.M.)$,the acceleration $a$ of a particle is directly proportional to the negative of its displacement $x$ from the mean position.
This relationship is given by the equation: $a = -\omega^2 x$,where $\omega$ is the angular frequency.
This equation is of the form $y = mx$,which represents a straight line passing through the origin.
Since the slope $m = -\omega^2$ is negative,the graph must be a straight line passing through the origin with a negative slope.
Therefore,the correct graph is the one shown in option $C$.

(D) The mass of the particle is $m = 10 \, g = 10 \times 10^{-3} \, kg = 0.01 \, kg$.
The amplitude is $A = 0.5 \, m$.
The angular frequency is $\omega = 10 \, rad/s$.
The maximum force $F_{max}$ in simple harmonic motion is given by the formula $F_{max} = m \omega^2 A$.
Substituting the values: $F_{max} = 0.01 \times (10)^2 \times 0.5$.
$F_{max} = 0.01 \times 100 \times 0.5$.
$F_{max} = 1 \times 0.5 = 0.5 \, N$.

(D) The acceleration $a$ of a particle in simple harmonic motion is given by $a = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the displacement from the equilibrium position.
Given $a = 12 \ cm/sec^2$ and $y = 3 \ cm$.
Substituting these values: $12 = \omega^2 \times 3$.
$\omega^2 = \frac{12}{3} = 4$.
$\omega = 2 \ rad/sec$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi \approx 3.14 \ sec$.

(D) The given equation is $y = \sin \left( \frac{\pi t}{4} + \frac{\pi}{6} \right)$.
Comparing this with the standard $SHM$ equation $y = A \sin(\omega t + \phi)$:
The amplitude $A = 1 \ cm$ and the angular frequency $\omega = \frac{\pi}{4} \ rad/sec$.
The maximum acceleration $a_{max}$ is given by the formula $a_{max} = \omega^2 A$.
Substituting the values: $a_{max} = \left( \frac{\pi}{4} \right)^2 \times 1$.
$a_{max} = \frac{\pi^2}{16} \approx \frac{9.8696}{16} \approx 0.6168 \ cm/sec^2$.
Rounding to two decimal places,we get $0.62 \ cm/sec^2$.

(C) Let the displacement of a particle executing simple harmonic motion be $y = A \sin(\omega t)$.
The instantaneous velocity $v$ is given by the derivative of displacement with respect to time:
$v = \frac{dy}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \frac{\pi}{2})$.
The instantaneous acceleration $a$ is given by the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi)$.
The phase of velocity is $(\omega t + \frac{\pi}{2})$ and the phase of acceleration is $(\omega t + \pi)$.
The phase difference between acceleration and velocity is $(\omega t + \pi) - (\omega t + \frac{\pi}{2}) = \frac{\pi}{2}$ or $0.5\pi$.

(D) The angular frequencies are given as $\omega_{1} = 100 \, rad \, s^{-1}$ and $\omega_{2} = 1000 \, rad \, s^{-1}$.
Let the common displacement amplitude be $A$.
The maximum acceleration $a_{max}$ for a simple harmonic motion is given by the formula $a_{max} = \omega^2 A$.
For the first motion,$a_{max1} = \omega_{1}^2 A = (100)^2 A$.
For the second motion,$a_{max2} = \omega_{2}^2 A = (1000)^2 A$.
The ratio of their maximum accelerations is $\frac{a_{max1}}{a_{max2}} = \frac{\omega_{1}^2 A}{\omega_{2}^2 A} = \frac{\omega_{1}^2}{\omega_{2}^2}$.
Substituting the values,we get $\frac{a_{max1}}{a_{max2}} = \frac{(100)^2}{(1000)^2} = \frac{10000}{1000000} = \frac{1}{100}$.
Thus,the ratio is $1:10^2$.

(C) Given the displacement equation: $x = A \cos \omega t$.
The velocity $v$ is the first derivative of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(A \cos \omega t) = -A \omega \sin \omega t$.
The acceleration $a$ is the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt}(-A \omega \sin \omega t) = -A \omega^2 \cos \omega t$.
Comparing this with the original displacement equation $x = A \cos \omega t$,we see that $a = -\omega^2 x$. At $t = 0$,$x = A$,so $a = -A \omega^2$. This means the acceleration starts at a negative maximum value and follows a negative cosine curve. Graph $C$ represents this variation correctly.

(C) The maximum acceleration is given by $a_{\max} = A\omega^2$ and the maximum speed is given by $v_{\max} = A\omega$.
Let the initial amplitude be $A_1$ and angular frequency be $\omega_1$. Then $v_{\max} = A_1\omega_1$ and $a_{\max} = A_1\omega_1^2$.
If the new amplitude is $A_2$ and new angular frequency is $\omega_2$,we are given that $v_{\max}$ remains constant,so $A_1\omega_1 = A_2\omega_2$.
We are also given that the new maximum acceleration is twice the original,so $A_2\omega_2^2 = 2(A_1\omega_1^2)$.
Dividing the second equation by the first: $\frac{A_2\omega_2^2}{A_1\omega_1} = \frac{2A_1\omega_1^2}{A_1\omega_1} \implies \frac{\omega_2}{\omega_1} = 2$,which means $\omega_2 = 2\omega_1$ (frequency is doubled).
Substituting $\omega_2 = 2\omega_1$ into $A_1\omega_1 = A_2\omega_2$,we get $A_1\omega_1 = A_2(2\omega_1) \implies A_2 = \frac{A_1}{2}$ (amplitude is halved).
Therefore,the frequency is doubled and the amplitude is halved.

(NONE) In simple harmonic motion $(SHM)$,the restoring force (or unbalanced force) $F$ acting on a particle is given by the equation $F = -kd$,where $k$ is the force constant and $d$ is the displacement from the equilibrium position.
This equation $F = -kd$ represents a linear relationship between the force $F$ and the displacement $d$,where the force is directly proportional to the displacement but acts in the opposite direction.
Graphically,this is a straight line passing through the origin with a negative slope.
None of the provided graphs $(I, II, III, IV)$ depict a straight line passing through the origin with a negative slope. Therefore,the question as presented is technically flawed because the correct graph for force versus displacement is not among the options.

(C) Given that the magnitude of velocity $|v|$ is equal to the magnitude of acceleration $|a|$.
For $SHM$,velocity is $v = \omega \sqrt{A^2 - x^2}$ and acceleration is $a = -\omega^2 x$.
Equating the magnitudes: $|v| = |a| \Rightarrow \omega \sqrt{A^2 - x^2} = \omega^2 x$.
Dividing both sides by $\omega$ (assuming $\omega \neq 0$): $\sqrt{A^2 - x^2} = \omega x$.
Solving for angular frequency $\omega$: $\omega = \frac{\sqrt{A^2 - x^2}}{x}$.
Given $A = 2 \, cm$ and $x = 1 \, cm$,we have $\omega = \frac{\sqrt{2^2 - 1^2}}{1} = \sqrt{3} \, rad/s$.
Since $\omega = 2 \pi f$,the frequency $f = \frac{\omega}{2 \pi} = \frac{\sqrt{3}}{2 \pi} \, s^{-1}$.

(A) The acceleration $a$ of a particle in $SHM$ is given by the equation $a = -\omega^{2}x$,where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.
Comparing this with the equation of a straight line $y = mx + c$,we see that $y = a$,$m = -\omega^{2}$,$x = x$,and $c = 0$.
Since the intercept $c$ is $0$,the graph is a straight line passing through the origin.
Since $\omega^{2}$ is always positive,the slope $m = -\omega^{2}$ is always negative.
Therefore,the graph is a straight line passing through the origin with a negative slope.

(A) Given the displacement equation: $x = x_0 \cos(\omega t - \frac{\pi}{4})$.
Velocity $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4})$.
Acceleration $a = \frac{dv}{dt} = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4})$.
Using the trigonometric identity $\cos(\theta + \pi) = -\cos(\theta)$,we can rewrite the acceleration as:
$a = x_0 \omega^2 \cos(\omega t - \frac{\pi}{4} + \pi) = x_0 \omega^2 \cos(\omega t + \frac{3\pi}{4})$.
Comparing this with the given form $a = A \cos(\omega t + \delta)$,we find:
$A = x_0 \omega^2$ and $\delta = \frac{3\pi}{4}$.

(B) For a particle executing Simple Harmonic Motion $(SHM)$,the displacement is given by $x = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
The acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x$.
From the acceleration equation,we have $\frac{a}{x} = -\omega^2$.
Since $\omega = \frac{2\pi}{T}$,we have $\omega^2 = \frac{4\pi^2}{T^2}$.
Substituting this into the ratio,we get $\frac{a}{x} = -\frac{4\pi^2}{T^2}$.
Rearranging this,we get $\frac{aT^2}{x} = -4\pi^2$,which is a constant.
However,looking at the options provided,we evaluate $\frac{a}{x} = -\omega^2$. Since $\omega$ and $T$ are constants for a given $SHM$,the ratio $\frac{a}{x}$ is constant. Thus,$\frac{aT}{x}$ is also a constant because $T$ is constant.

(D) For a particle in simple harmonic motion, the acceleration $A$ is given by $A = -\omega^2 x$, where $\omega$ is the angular frequency.
Taking the magnitude, we have $|A| = \omega^2 |x|$, which implies $\omega^2 = \frac{|A|}{|x|}$.
Using the values from the table (e.g., $|A| = 16 \ mm \ s^{-2}$ and $|x| = 4 \ mm$):
$\omega^2 = \frac{16 \ mm \ s^{-2}}{4 \ mm} = 4 \ s^{-2}$.
Therefore, $\omega = \sqrt{4} = 2 \ rad \ s^{-1}$.
The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2\pi}{2} = \pi \ s$.

(B) The displacement $y$ of a particle in $SHM$ is given by $y = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Acceleration $a$ is the second derivative of displacement with respect to time:
$a = \frac{d^2y}{dt^2} = \frac{d}{dt}(A \omega \cos(\omega t)) = -A \omega^2 \sin(\omega t)$.
Comparing the equations,$a = -\omega^2 y$.
This means the acceleration graph is an inverted version of the displacement graph,scaled by a factor of $\omega^2$.
Since the displacement graph starts at $0$ and goes positive,the acceleration graph must start at $0$ and go negative.
Looking at the options,option $B$ shows a graph that starts at $0$ and goes negative,which matches the derived equation $a = -A \omega^2 \sin(\omega t)$.

(A) At the instant indicated by the dashed line,the particle is at its negative extreme position $(x = -A)$.
At the extreme position,the velocity of the particle is zero $(v = 0)$.
The restoring force in a spring is given by $F = -kx$. Since the particle is at $x = -A$,the force is $F = -k(-A) = +kA$.
$A$ positive force indicates that the force is directed towards the positive direction (to the right).
Therefore,the velocity is zero and the force is to the right.

(A) For a particle executing $SHM$,the acceleration $a$ is given by $a = -\omega^{2}x$.
The slope of the acceleration-displacement graph is $m = \frac{a}{x} = -\omega^{2}$.
From the given graph,the angle made by the line with the negative $x$-axis is $37^{\circ}$. The slope of the line is $\tan(180^{\circ} - 37^{\circ}) = -\tan 37^{\circ} = -\frac{3}{4}$.
Equating the slopes: $-\omega^{2} = -\frac{3}{4} \Rightarrow \omega^{2} = \frac{3}{4}$.
Since $\omega = \frac{2\pi}{T}$,we have $\omega^{2} = \frac{4\pi^{2}}{T^{2}} = \frac{3}{4}$.
Solving for $T$: $T^{2} = \frac{16\pi^{2}}{3} \Rightarrow T = \frac{4\pi}{\sqrt{3}} \, s$.

(B) Maximum velocity in $SHM$ is given by $v_{\max} = a\omega$.
Maximum acceleration in $SHM$ is given by $A_{\max} = a\omega^2$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Given the ratio $\frac{A_{\max}}{v_{\max}} = 10$,we have $\frac{a\omega^2}{a\omega} = 10$,which implies $\omega = 10\,s^{-1}$.
The displacement equation is $x = a \sin(\omega t + \phi)$.
At $t = 0$,$x = 5\,m$ and $\phi = \frac{\pi}{4}$.
Substituting these values: $5 = a \sin(\frac{\pi}{4}) = a \cdot \frac{1}{\sqrt{2}}$.
Thus,the amplitude $a = 5\sqrt{2}\,m$.
The maximum acceleration is $A_{\max} = a\omega^2 = (5\sqrt{2}) \cdot (10)^2 = 5\sqrt{2} \cdot 100 = 500\sqrt{2}\,m/s^2$.

(B) For a particle executing simple harmonic motion,the force is given by $F = -kx$,where $k$ is the force constant.
From the graph,at $x = 20\, cm = 0.2\, m$,the force $F = -2.0\, N$.
Thus,the magnitude of the force constant is $k = \frac{|F|}{|x|} = \frac{2.0\, N}{0.2\, m} = 10\, N/m$.
The mass of the particle is $m = 400\, g = 0.4\, kg$.
The frequency of oscillation $f$ is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
Substituting the values,we get $f = \frac{1}{2\pi} \sqrt{\frac{10}{0.4}} = \frac{1}{2\pi} \sqrt{25} = \frac{5}{2\pi}\, s^{-1}$.

(A) For a particle in simple harmonic motion,the maximum acceleration is given by $a_{\max} = \omega^2 A = \alpha$ $...(1)$
The maximum velocity is given by $v_{\max} = \omega A = \beta$ $...(2)$
Dividing equation $(1)$ by equation $(2)$,we get:
$\frac{\omega^2 A}{\omega A} = \frac{\alpha}{\beta}$
$\omega = \frac{\alpha}{\beta}$
Since angular frequency $\omega = 2\pi f$,where $f$ is the frequency of vibration:
$2\pi f = \frac{\alpha}{\beta}$
$f = \frac{\alpha}{2\pi\beta}$

(B) In $SHM$,the acceleration $\vec a$ is given by $\vec a = -\omega^2 \vec r$,where $\vec r$ is the displacement vector from the equilibrium position.
Taking the dot product of $\vec a$ and $\vec r$:
$\vec a \cdot \vec r = (-\omega^2 \vec r) \cdot \vec r = -\omega^2 |\vec r|^2$.
Since $\omega^2$ is always positive and $|\vec r|^2$ is always non-negative,the product $\vec a \cdot \vec r$ is always negative (except at the equilibrium position where it is zero).
Therefore,the correct option is $B$.

(A) Given the displacement equation: $x = x_0 \cos(\omega t + \pi/4)$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t + \pi/4)$.
The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -x_0 \omega^2 \cos(\omega t + \pi/4)$.
Using the trigonometric identity $-\cos(\theta) = \cos(\theta + \pi)$,we can rewrite the acceleration as:
$a = x_0 \omega^2 \cos(\omega t + \pi/4 + \pi) = x_0 \omega^2 \cos(\omega t + 5\pi/4)$.
Since $\cos(\theta)$ has a period of $2\pi$,$\cos(\omega t + 5\pi/4)$ is equivalent to $\cos(\omega t + 5\pi/4 - 2\pi) = \cos(\omega t - 3\pi/4)$.
Comparing $a = x_0 \omega^2 \cos(\omega t - 3\pi/4)$ with $a = A \cos(\omega t + \delta)$,we get $A = x_0 \omega^2$ and $\delta = -3\pi/4$ or $5\pi/4$.
Note: Checking the provided options,the correct phase shift corresponds to $3\pi/4$ if the initial phase was $-\pi/4$. Given the input equation $x = x_0 \cos(\omega t + \pi/4)$,the acceleration is $a = -x_0 \omega^2 \cos(\omega t + \pi/4) = x_0 \omega^2 \cos(\omega t + \pi/4 + \pi) = x_0 \omega^2 \cos(\omega t + 5\pi/4)$. This is equivalent to $x_0 \omega^2 \cos(\omega t - 3\pi/4)$. If the question intended $x = x_0 \cos(\omega t - \pi/4)$,then $a = x_0 \omega^2 \cos(\omega t - \pi/4 + \pi) = x_0 \omega^2 \cos(\omega t + 3\pi/4)$. Thus,$A = x_0 \omega^2$ and $\delta = 3\pi/4$.

(B) Let $m$ be the mass of the coin and $N$ be the normal reaction force exerted by the platform on the coin.
The equation of motion for the coin in the vertical direction is given by:
$mg - N = m a$
where $a$ is the acceleration of the platform. For simple harmonic motion,the acceleration is $a = -\omega^2 x$,where $x$ is the displacement from the mean position.
Substituting this into the equation of motion:
$mg - N = m(-\omega^2 x)$
$N = m(g + \omega^2 x)$
However,when the platform moves upwards,the acceleration is directed downwards,so $a = -\omega^2 x$ (where $x$ is positive upwards). When the platform is at the highest point,$x = A$ (amplitude),and the acceleration is $a = -\omega^2 A$.
The equation becomes $mg - N = m(-\omega^2 A)$,which implies $N = m(g + \omega^2 A)$. This does not lead to $N=0$.
When the platform moves downwards,the acceleration is directed upwards. At the highest point of the oscillation,the platform's acceleration is directed downwards. The coin loses contact when the normal force $N$ becomes zero.
This happens when the downward acceleration of the platform exceeds the acceleration due to gravity $g$.
The maximum downward acceleration is $\omega^2 A$. Thus,the coin leaves contact when $\omega^2 A \ge g$,or $A \ge \frac{g}{\omega^2}$.
Therefore,the coin will leave contact for the first time when the amplitude reaches $\frac{g}{\omega^2}$ at the highest position of the platform.

(B) The given equation for $S.H.M.$ is $y = A \sin(\omega t + \phi)$,where $A = 2 \, cm$ and $\omega = \frac{\pi}{2} \, rad/s$.
The velocity $v$ is given by $\frac{dy}{dt} = A\omega \cos(\omega t + \phi)$.
The acceleration $a$ is given by $\frac{d^2y}{dt^2} = -A\omega^2 \sin(\omega t + \phi)$.
The maximum acceleration $a_{max}$ is given by $|A\omega^2|$.
Substituting the values: $a_{max} = 2 \times \left( \frac{\pi}{2} \right)^2 = 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} \, cm/s^2$.