Acceleration and Force of Simple Harmonic Motion Questions in English

(A) $SHM$ is defined as a to-and-fro motion about a mean position. The restoring force,and consequently the acceleration,is always directed towards the mean position to bring the body back. The acceleration in $SHM$ is given by $a = -\omega^2 x$,where $x$ is the displacement from the mean position. Since the negative sign indicates that the acceleration is always opposite to the direction of displacement,it is always directed towards the mean position. The reason provided correctly explains that the body must stop at the extreme position and return to the mean position due to this acceleration. Therefore,both statements are correct and the reason explains the assertion.

(N/A) The mean position $O$ is the midpoint of $AB$. $A$ is at $x = -5 \; cm$ and $B$ is at $x = +5 \; cm$. The positive direction is $A \to B$. In $SHM$,acceleration $a = -\omega^2 x$ and force $F = ma = -m\omega^2 x$.
$(a)$ At $A$ $(x = -5 \; cm)$: Velocity $v = 0$ (extreme point). Acceleration $a = -\omega^2(-5) > 0$ (positive). Force $F$ is positive.
$(b)$ At $B$ $(x = +5 \; cm)$: Velocity $v = 0$ (extreme point). Acceleration $a = -\omega^2(+5) < 0$ (negative). Force $F$ is negative.
$(c)$ At $O$ $(x = 0)$ going towards $A$: Velocity $v < 0$ (moving left). Acceleration $a = 0$. Force $F = 0$.
$(d)$ At $2 \; cm$ from $B$ towards $A$ $(x = +3 \; cm)$: Velocity $v < 0$ (moving left). Acceleration $a = -\omega^2(+3) < 0$ (negative). Force $F$ is negative.
$(e)$ At $3 \; cm$ from $A$ towards $B$ $(x = -2 \; cm)$: Velocity $v > 0$ (moving right). Acceleration $a = -\omega^2(-2) > 0$ (positive). Force $F$ is positive.
$(f)$ At $4 \; cm$ from $B$ towards $A$ $(x = +1 \; cm)$: Velocity $v < 0$ (moving left). Acceleration $a = -\omega^2(+1) < 0$ (negative). Force $F$ is negative.

(N/A) $1$. Consider a particle $P$ moving in a circle of radius $A$ with constant angular velocity $\omega$. The projection of this motion on the diameter (say $x$-axis) represents $SHM$.
$2$. The position of the particle at any time $t$ is given by $x = A \cos(\omega t + \phi)$.
$3$. The velocity of the particle is $v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$.
$4$. The instantaneous acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-A\omega \sin(\omega t + \phi))$.
$5$. Differentiating,we get $a = -A\omega^2 \cos(\omega t + \phi)$.
$6$. Since $x = A \cos(\omega t + \phi)$,we can substitute this into the acceleration equation to get $a = -\omega^2 x$.

(N/A) The acceleration of an $SHM$ particle is the second derivative of displacement with respect to time.
The displacement of an $SHM$ particle at time $t$ is given by:
$x(t) = A \cos(\omega t + \phi)$
Differentiating with respect to time $t$ to find velocity $v(t)$:
$v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)$
Differentiating again with respect to time $t$ to find acceleration $a(t)$:
$a(t) = \frac{dv}{dt} = -A \omega^2 \cos(\omega t + \phi)$
Since $x(t) = A \cos(\omega t + \phi)$,we can substitute this into the acceleration equation:
$a(t) = -\omega^2 x(t)$
Generally,$a = -\omega^2 x$.
Special cases:
$(1)$ At the mean position,$x = 0$,so $a = -\omega^2(0) = 0$. Acceleration is zero and velocity is maximum.
$(2)$ At extreme points,$x = \pm A$,so $a = \mp \omega^2 A$. The magnitude of acceleration is maximum,$a_{\max} = A \omega^2$.

(B) For a particle executing $SHM$,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 $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x$.
At the extreme positions,the displacement $x = \pm A$.
Therefore,the velocity $v = \pm A\omega \cos(\pm \pi/2) = 0$.
The acceleration $a = -\omega^2(\pm A) = \mp A\omega^2$,which is the maximum magnitude of acceleration.
Thus,both conditions are satisfied at the extreme positions.

(N/A) For a particle executing $SHM$ along the $X$-axis,the displacement $x$ at any time $t$ is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
The instantaneous acceleration $a$ is the second derivative of displacement with respect to time:
$a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = -A\omega^2 \sin(\omega t + \phi)$.
Since $x = A \sin(\omega t + \phi)$,we can substitute this into the expression for acceleration:
$a = -\omega^2 x$.
Thus,the formula for instantaneous acceleration is $a = -\omega^2 x$.

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

(B) In $SHM$ (Simple Harmonic Motion),the particle moves back and forth about a mean position.
$1$. Velocity: The velocity of the particle is always directed along the instantaneous direction of its motion.
$2$. Acceleration: The acceleration in $SHM$ is given by $a = -\omega^2 x$,where $x$ is the displacement from the mean position. The negative sign indicates that the acceleration is always directed towards the mean position,regardless of the direction of motion or displacement.

(D) Displacement of $SHM$ particle at time $t$ is given by:
$x(t) = A \cos (\omega t + \phi)$
where $A$ is the amplitude,$\omega$ is the angular frequency,and $\phi$ is the initial phase.
Differentiating the displacement equation with respect to time $t$ to find velocity:
$v(t) = \frac{d}{dt} [A \cos (\omega t + \phi)] = -A\omega \sin (\omega t + \phi)$
Differentiating the velocity equation with respect to time $t$ to find acceleration:
$a(t) = \frac{d}{dt} [-A\omega \sin (\omega t + \phi)] = -A\omega^2 \cos (\omega t + \phi)$
Since $x(t) = A \cos (\omega t + \phi)$,we can substitute this into the acceleration equation:
$a(t) = -\omega^2 x(t)$
According to Newton's second law,$F = ma$. Multiplying both sides by mass $m$:
$F = m a(t) = -m\omega^2 x(t)$
Defining the force constant $k = m\omega^2$,we get:
$F = -kx(t)$
This shows that the restoring force is directly proportional to the negative displacement,which is the force law for $SHM$.

(N/A) In Simple Harmonic Motion $(SHM)$,the restoring force $F$ acting on a particle is directly proportional to its displacement $x$ from the equilibrium position and is always directed towards the equilibrium position.
Mathematically,this is expressed as:
$F = -kx$
Where:
$F$ is the restoring force.
$k$ is the force constant (or spring constant),which is a positive constant.
$x$ is the displacement from the mean position.
The negative sign indicates that the force is always in the direction opposite to the displacement.

(A) For a particle of mass $m$ executing Simple Harmonic Motion $(SHM)$,the restoring force is given by $F = -kx$,where $k$ is the force constant and $x$ is the displacement.
From the equation of motion for $SHM$,we have $F = ma = m\frac{d^2x}{dt^2}$.
Since $a = -\omega^2x$,we can write $F = m(-\omega^2x) = -(m\omega^2)x$.
Comparing $F = -kx$ with $F = -(m\omega^2)x$,we get $k = m\omega^2$.

(N/A) restoring force is a force that acts to bring a body to its equilibrium position whenever it is displaced from that position.
Mathematically,for a linear system like a spring,it is given by Hooke's Law: $F = -kx$,where $F$ is the restoring force,$k$ is the force constant,and $x$ is the displacement from the equilibrium position.
The negative sign indicates that the force is always directed opposite to the direction of displacement.

(B) Let the displacement of a particle executing $SHM$ be $x = A \sin(\omega t + \phi)$.
The velocity $v$ is given by $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) = A \omega \sin(\omega t + \phi + \frac{\pi}{2})$.
The acceleration $a$ is given by $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = A \omega^2 \sin(\omega t + \phi + \pi)$.
The phase of velocity is $(\omega t + \phi + \frac{\pi}{2})$ and the phase of acceleration is $(\omega t + \phi + \pi)$.
The phase difference is $(\omega t + \phi + \pi) - (\omega t + \phi + \frac{\pi}{2}) = \frac{\pi}{2}$ radians or $90^{\circ}$.

(C) For a particle executing $SHM$,the displacement $y$ is given by $y = A \sin(\omega t + \phi)$.
The acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{d^2y}{dt^2} = -\omega^2 A \sin(\omega t + \phi)$.
We can rewrite this as $a = \omega^2 A \sin(\omega t + \phi + \pi)$.
Comparing the arguments of the sine functions,the phase of displacement is $(\omega t + \phi)$ and the phase of acceleration is $(\omega t + \phi + \pi)$.
The phase difference is $(\omega t + \phi + \pi) - (\omega t + \phi) = \pi \text{ radians}$ or $180^{\circ}$.

(B) The restoring force in $SHM$ is given by $F = -kx$,where $k$ is the force constant and $x$ is the displacement from the mean position.
At the extreme position,the displacement $x = A = 2 \, cm = 0.02 \, m$.
The force at the extreme position is $F_{ext} = kA = 4 \, N$.
Therefore,$k = \frac{4}{0.02} = 200 \, N/m$.
At the midway point between the mean position $(x = 0)$ and the extreme point $(x = A)$,the displacement is $x = \frac{A}{2} = \frac{2 \, cm}{2} = 1 \, cm = 0.01 \, m$.
The force at this point is $F = kx = 200 \times 0.01 = 2 \, N$.

(B) In simple harmonic motion $(SHM)$, the restoring force acting on a particle is directly proportional to its displacement from the equilibrium position and is directed towards the equilibrium position.
Mathematically, this is expressed as $F = -kx$, where $F$ is the restoring force, $x$ is the displacement from the mean position, and $k$ is the force constant (or spring constant).

(B) The displacement $(x)$ of a particle in $SHM$ is given by $x = A \sin(\omega t + \phi)$.
The velocity $(v)$ is the first derivative of displacement: $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.
The acceleration $(a)$ is the second derivative of displacement: $a = \frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t + \phi)$.
Using the trigonometric identity $-\sin(\theta) = \sin(\theta + \pi)$,we can rewrite the acceleration as $a = A \omega^2 \sin(\omega t + \phi + \pi)$.
Comparing the phase of displacement $(\omega t + \phi)$ and acceleration $(\omega t + \phi + \pi)$,the phase difference is $\pi \; rad$.

(D) The magnitude of acceleration in simple harmonic motion is given by the formula $|a| = \omega^2 y$,where $a$ is acceleration,$\omega$ is angular frequency,and $y$ is displacement.
Given $a = 1.0 \, m/s^2$ and $y = 0.01 \, m$,we have:
$1.0 = \omega^2 \times 0.01$
$\omega^2 = \frac{1.0}{0.01} = 100$
$\omega = 10 \, rad/s$
The time period $T$ is related to angular frequency by $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$:
$T = \frac{2\pi}{10} = \frac{\pi}{5} \, s$.

(B) The book will lose contact with the shelf when the downward acceleration of the shelf exceeds the acceleration due to gravity $g$.
The condition for losing contact is $a_{\max} \geq g$.
Since the shelf undergoes simple harmonic motion,the maximum acceleration is given by $a_{\max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Setting $a_{\max} = g$,we get $\omega^2 A = g$,which implies $\omega = \sqrt{\frac{g}{A}}$.
Given $A = 2.5 \,cm = 0.025 \,m$ and $g = 10 \,m/s^2$,we have $\omega = \sqrt{\frac{10}{0.025}} = \sqrt{400} = 20 \,rad/s$.
The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{20}{2\pi} = \frac{10}{\pi} \approx 3.18 \,Hz$.

(C) For a particle executing simple harmonic motion $(S.H.M.)$,the acceleration $a$ is given by the relation:
$a = -\omega^2 x$
where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.
Since the acceleration $a$ is directly proportional to the displacement $x$,it changes as the particle moves.
Therefore,the acceleration is non-uniform because it depends on the position $x$ of the particle,not directly on time in a linear fashion (it varies sinusoidally with time).
Thus,the correct option is $(c)$.

(D) Given: Mass $m = 10 \,g = 0.01 \,kg$,Amplitude $A = 10 \,cm = 0.1 \,m$,Period $T = 0.1 \,s$.
The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.1} = 20\pi \,rad/s$.
The maximum acceleration in $S.H.M.$ is $a_{\text{max}} = \omega^2 A$.
The maximum force is $F_{\text{max}} = m \cdot a_{\text{max}} = m \omega^2 A$.
Substituting the values: $F_{\text{max}} = 0.01 \times (20\pi)^2 \times 0.1$.
$F_{\text{max}} = 0.001 \times 400 \pi^2 = 0.4 \pi^2$.
Using $\pi^2 \approx 9.87$,$F_{\text{max}} \approx 0.4 \times 9.87 = 3.948 \,N$.
Rounding to the nearest integer,$F_{\text{max}} \approx 4 \,N$.

(B) The equation of motion is $x = A \cos(\omega t + \phi)$,where $A = 5.0 \, m$ and $\omega = 2 \pi \, rad \, s^{-1}$.
The acceleration $a$ in $S.H.M.$ is given by $a = -\omega^2 x$.
First,find the displacement $x$ at $t = 1.5 \, s$:
$x = 5 \cos(2 \pi \times 1.5 + \pi / 4) = 5 \cos(3 \pi + \pi / 4)$.
Since $\cos(3 \pi + \theta) = -\cos(\theta)$:
$x = -5 \cos(\pi / 4) = -5 / \sqrt{2} \, m$.
Now,calculate the acceleration:
$a = -\omega^2 x = -(2 \pi)^2 \times (-5 / \sqrt{2})$.
$a = 4 \pi^2 \times (5 / \sqrt{2}) = 4 \times 9.8696 \times 3.5355 \approx 139.56 \, m / s^2$.

(A) From the graph,the amplitude $A = 1 \ m$ and the time period $T = 8 \ s$.
The angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \ rad/s$.
The equation of motion is $x(t) = A \sin(\omega t) = 1 \sin\left(\frac{\pi}{4} t\right)$.
The acceleration $a(t)$ is given by $a = \frac{d^2x}{dt^2} = -\omega^2 x = -\omega^2 A \sin(\omega t)$.
Substituting the values at $t = 2 \ s$:
$a = -\left(\frac{\pi}{4}\right)^2 \times 1 \times \sin\left(\frac{\pi}{4} \times 2\right)$
$a = -\frac{\pi^2}{16} \times \sin\left(\frac{\pi}{2}\right)$
Since $\sin\left(\frac{\pi}{2}\right) = 1$,we get:
$a = -\frac{\pi^2}{16} \ m/s^2$.

(D) The correct option is $D$.
From the $x-t$ graph, at $t = 0$, the particle is at $x = 0$.
Hence, the particle starts its simple harmonic motion $(SHM)$ from the mean position.
The general equation for $SHM$ is $x = A \sin(\omega t)$.
From the graph, the time period $T = 8 \,s$ and the amplitude $A = 1 \,cm$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \,rad/s$.
Thus, the equation of motion is $x = 1 \sin\left(\frac{\pi t}{4}\right)$.
At $t = 4/3 \,s$, the displacement is $x = \sin\left(\frac{\pi}{4} \times \frac{4}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \,cm$.
The acceleration $a$ is given by $a = -\omega^2 x$.
Substituting the values, $a = -\left(\frac{\pi}{4}\right)^2 \times \frac{\sqrt{3}}{2} = -\frac{\pi^2}{16} \times \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}\pi^2}{32} \,cm/s^2$.

(D) The total energy $E$ in $S.H.M.$ is given by $E = \frac{1}{2} k A^2$, where $k = m \omega^2$ is the force constant and $A$ is the amplitude.
Given that the kinetic energy $K$ and potential energy $U$ are equal, $K = U = \frac{E}{2}$.
The potential energy at displacement $x$ is $U = \frac{1}{2} k x^2$.
Thus, $\frac{1}{2} k x^2 = \frac{1}{2} (\frac{1}{2} k A^2) \implies x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}$.
The force acting on the particle at displacement $x$ is $F = -kx = -m \omega^2 x$.
The maximum force is $F_m = k A = 50 \,N$.
The magnitude of the force at displacement $x$ is $F' = kx = k (\frac{A}{\sqrt{2}}) = \frac{F_m}{\sqrt{2}}$.
Substituting $F_m = 50 \,N$, we get $F' = \frac{50}{\sqrt{2}} = 25 \sqrt{2} \,N$.

(C) For a particle performing Simple Harmonic Motion ($S$.$H$.$M$.),the potential energy $E$ at displacement $x$ is given by $E = \frac{1}{2} k x^2$,where $k$ is the force constant.
The restoring force $F$ acting on the particle at displacement $x$ is given by $F = -k x$.
From the expression for potential energy,we have $2 E = k x^2$.
Dividing this equation by the expression for force $F = -k x$,we get:
$\frac{2 E}{F} = \frac{k x^2}{-k x}$
Simplifying the expression,we get:
$\frac{2 E}{F} = -x$
Rearranging the terms,we obtain:
$\frac{2 E}{F} + x = 0$
Thus,the correct relation is $\frac{2 E}{F} + x = 0$.

(A) In $S.H.M.$, the displacement is given by $y = A \sin(\omega t)$.
The force acting on the particle is $F = -ky = -k A \sin(\omega t)$, where $k$ is the force constant.
This equation shows that the force is $180^{\circ}$ (or $\pi$ radians) out of phase with the displacement.
If the displacement graph is a sine wave starting from the origin, the force graph must be an inverted sine wave (a negative sine wave) starting from the origin.
Therefore, the force-time graph will be the mirror image of the displacement-time graph about the time axis.

(B) The maximum acceleration $a_{max}$ of a particle executing simple harmonic motion is given by the formula $a_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.
Given that the amplitudes $A_1$ and $A_2$ are equal $(A_1 = A_2 = A)$,the ratio of the maximum accelerations is:
$\frac{a_{max,1}}{a_{max,2}} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left( \frac{\omega_1}{\omega_2} \right)^2$
Substituting the given values $\omega_1 = 300 \ rad/s$ and $\omega_2 = 3000 \ rad/s$:
$\frac{a_{max,1}}{a_{max,2}} = \left( \frac{300}{3000} \right)^2 = \left( \frac{1}{10} \right)^2 = \frac{1}{100} = 1: 10^2$
Therefore,the correct option is $B$.

(A) In $S.H.M.$,the displacement of a particle is given by $x(t) = A \cos(\omega t + \phi)$.
Since the particle starts from the extreme position,at $t = 0$,$x = A$,which implies $\phi = 0$. Thus,$x(t) = A \cos(\omega t)$.
The acceleration $a(t)$ is given by the second derivative of displacement: $a(t) = \frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t)$.
We can rewrite the acceleration as $a(t) = \omega^2 A \cos(\omega t + \pi)$.
Comparing the phase of displacement $(\omega t)$ and acceleration $(\omega t + \pi)$,the phase difference is $\pi \ rad$.

(B) The correct option is $(B)$.
Concept: For $SHM$,the acceleration is given by $a = -\omega^2 x$.
The displacement is $x = A \sin(\omega t + \phi)$.
Velocity is the first derivative of displacement: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
Acceleration is the derivative of velocity: $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t + \phi)$.
At the extreme positions,displacement $x = \pm A$,so acceleration $a = \mp A\omega^2$ is maximum (in magnitude),and velocity $v = 0$.
At the mean position,displacement $x = 0$,so acceleration $a = 0$,and velocity $v = \pm A\omega$ is maximum.
Therefore,when '$a$' is maximum,'$v$' is zero.

(C) Concept: The wooden cube will leave the plank at the top extreme position if the downward acceleration of the plank exceeds the acceleration due to gravity.
For the block not to leave the plank,the normal force $N$ must be greater than or equal to zero.
From the free body diagram of the block,the equation of motion is: $mg - N = ma$.
For the block to remain in contact,$N \geq 0$,which implies $mg \geq ma$,or $a \leq g$.
In $S$.$H$.$M$.,the maximum acceleration is given by $a_{max} = \omega^2 A$.
Therefore,the condition for the block not to leave the plank is $\omega^2 A \leq g$,or $A \leq \frac{g}{\omega^2}$.
Given frequency $f = \frac{3}{\pi} \text{ Hz}$.
Angular frequency $\omega = 2\pi f = 2\pi \left( \frac{3}{\pi} \right) = 6 \text{ rad/s}$.
Substituting the values: $A \leq \frac{10}{6^2} = \frac{10}{36} = \frac{5}{18} \text{ m}$.
Thus,the maximum amplitude is $\frac{5}{18} \text{ m}$.

(A) Given: Mass $m = 5 \,g = 5 \times 10^{-3} \,kg$,Amplitude $A = 0.3 \,m$,Time period $T = \frac{\pi}{5} \,s$.
Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi/5} = 10 \,rad/s$.
The maximum force acting on a particle in $S.H.M.$ is given by $F_{max} = m\omega^2A$.
Substituting the values: $F_{max} = (5 \times 10^{-3} \,kg) \times (10 \,rad/s)^2 \times (0.3 \,m)$.
$F_{max} = 5 \times 10^{-3} \times 100 \times 0.3 = 5 \times 10^{-1} \times 0.3 = 0.5 \times 0.3 = 0.15 \,N$.

(C) The given equation for displacement is $x = 5 \sin (3t + 3)$.
Comparing this with the standard equation of $S.H.M.$,$x = A \sin (\omega t + \phi)$,we get:
Amplitude $A = 5 \ cm$
Angular frequency $\omega = 3 \ rad/s$
The formula for the maximum acceleration of a particle in $S.H.M.$ is $a_{max} = A \omega^2$.
Substituting the values,we get $a_{max} = 5 \times (3)^2$.
$a_{max} = 5 \times 9 = 45 \ cm \ s^{-2}$.

(A) In simple harmonic motion,the acceleration $a$ is given by the formula $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.
At the mean position,the displacement $x = 0$.
Substituting $x = 0$ into the formula,we get $a = -\omega^2 (0) = 0$.
Therefore,the acceleration of the weight is zero at the mean position.

(A) For a particle performing $S.H.M.$,the displacement $x$ as a function of time $t$ is given by $x = A \sin(\omega t)$,where $A$ is the amplitude and $\omega$ is the angular frequency.
According to Hooke's Law for $S.H.M.$,the restoring force $F$ is given by $F = -kx$,where $k$ is the force constant.
Substituting the expression for $x$,we get $F = -k(A \sin(\omega t)) = -kA \sin(\omega t)$.
This equation shows that the force $F$ is proportional to the negative of the displacement $x$.
Therefore,the force-time graph will be an inverted version of the displacement-time graph. If the displacement graph starts from the origin and goes positive,the force graph must start from the origin and go negative.

(D) The span of oscillation $L$ is the total distance between the extreme positions,which is equal to $2A$,where $A$ is the amplitude of oscillation.
Therefore,$A = \frac{L}{2}$.
The angular frequency $\omega$ is given by $\omega = 2 \pi f$.
The acceleration $a$ of a particle in $S.H.M.$ is given by $a = -\omega^{2} x$.
At the top of the oscillation (extreme position),the displacement $x = A = \frac{L}{2}$.
Substituting the values,the magnitude of acceleration is $|a| = \omega^{2} A = (2 \pi f)^{2} \times \frac{L}{2}$.
$|a| = 4 \pi^{2} f^{2} \times \frac{L}{2} = 2 \pi^{2} f^{2} L$.
Thus,the correct option is $D$.

(D) The path length of $SHM$ is equal to $2A$,where $A$ is the amplitude.
Given path length $= 0.01 \ m$,so $2A = 0.01 \ m$,which implies $A = 0.005 \ m$.
The frequency $f = 50 \ Hz$.
The maximum force $F_{max}$ acting on a particle in $SHM$ is given by $F_{max} = m \omega^2 A$.
Since $\omega = 2 \pi f$,we have $F_{max} = m (2 \pi f)^2 A = m (4 \pi^2 f^2) A$.
Substituting the values: $F_{max} = 1 \times 4 \times \pi^2 \times (50)^2 \times 0.005$.
$F_{max} = 4 \times \pi^2 \times 2500 \times 0.005$.
$F_{max} = 10000 \times \pi^2 \times 0.005 = 50 \pi^2 \ N$.

(C) The acceleration of a particle performing $SHM$ is given by $a = -\omega^{2} x$,where $x = A \sin(\omega t + \phi)$.
To find the average acceleration over one complete oscillation (time period $T$),we integrate the acceleration over the interval $[0, T]$ and divide by $T$.
$\text{Average acceleration} = \frac{1}{T} \int_{0}^{T} a(t) dt = \frac{1}{T} \int_{0}^{T} -\omega^{2} A \sin(\omega t + \phi) dt$.
Since the integral of a sine function over one full period is zero,the average acceleration is $0$.

(B) The maximum velocity of a particle in $SHM$ is given by $v = a \omega$,where $a$ is the amplitude of displacement and $\omega$ is the angular frequency.
From this,we can find the angular frequency: $\omega = \frac{v}{a}$.
The maximum acceleration (amplitude of acceleration) is given by $A_{max} = \omega^2 a$.
Substituting the value of $\omega$ into the acceleration formula:
$A_{max} = \left(\frac{v}{a}\right)^2 \times a = \frac{v^2}{a^2} \times a = \frac{v^2}{a}$.

(C) For the object to remain on the platform, the normal force $N$ must be greater than or equal to zero. The equation of motion for the object of mass $m$ is given by $mg - N = ma$, where $a$ is the acceleration of the platform.
For the object to just lose contact with the surface, the normal force $N$ becomes $0$.
Thus, $mg = ma$, which implies $a = g$.
The acceleration of a particle in $S.H.M.$ is given by $a = A\omega^2$, where $A$ is the amplitude and $\omega$ is the angular frequency.
To avoid detachment, the maximum downward acceleration of the platform must not exceed $g$.
Therefore, $A\omega^2 \leq g$.
Substituting the given values, $A = 40 \text{ cm} = 0.4 \text{ m}$ and $g = 10 \text{ m/s}^2$:
$0.4 \omega^2 = 10$
$\omega^2 = \frac{10}{0.4} = 25$
$\omega = 5 \text{ rad/s}$.
Since $\omega = \frac{2\pi}{T}$, we have:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{5} = 0.4\pi \text{ s}$.
Thus, the least period of oscillation is $0.4\pi \text{ s}$.

(B) Given frequency $f = 0.5 \,Hz$.
Angular frequency $\omega = 2 \pi f = 2 \pi (0.5) = \pi \,rad/s$.
For the block to remain in contact with the piston, the downward acceleration of the piston at the extreme position must not exceed the acceleration due to gravity $g$.
The condition for the block to just lose contact is $a_{max} = g$.
Since $a_{max} = \omega^2 A$, we have $\omega^2 A = g$.
Substituting the values, $\pi^2 A = 10$.
Taking $\pi^2 \approx 10$, we get $10 A = 10$.
Therefore, $A = 1 \,m$.

(D) Given the force equation $F = -75 y$. Comparing this with the standard $SHM$ force equation $F = -ky$, we get the force constant $k = 75 \,N/m$.
Given mass $m = 3 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{75}{3}} = \sqrt{25} = 5 \,rad/s$.
The velocity at the mean position is the maximum velocity $V_{\max} = 2.5 \,m/s$.
Since $V_{\max} = A\omega$, the amplitude $A = \frac{V_{\max}}{\omega} = \frac{2.5}{5} = 0.5 \,m$.
The maximum acceleration is given by $a_{\max} = \omega^2 A$.
Substituting the values, $a_{\max} = (5)^2 \times 0.5 = 25 \times 0.5 = 12.5 \,m/s^2$.

(D) The displacement of the particle is given by $x = x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$.
Acceleration $a$ is given by $a = \frac{d^2x}{dt^2} = -\omega^2 x$.
Substituting the expression for $x$:
$a = -\omega^2 x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$.
Using the trigonometric identity $-\sin(\theta) = \sin(\theta + \pi)$:
$a = \omega^2 x_0 \sin \left(\omega t - \frac{\pi}{6} + \pi\right)$.
$a = \omega^2 x_0 \sin \left(\omega t + \frac{5\pi}{6}\right)$.
Comparing this with $a = A \sin(\omega t + \delta)$,we get $A = x_0 \omega^2$ and $\delta = \frac{5\pi}{6}$.

(C) The maximum force on a particle performing simple harmonic motion $(SHM)$ is given by $F_{\max} = m \omega^2 a = 10 \,N$, where $m$ is the mass, $\omega$ is the angular frequency, and $a$ is the amplitude.
At the mean position, the displacement $y = 0$. At the extreme position, the displacement $y = a$.
The force on the particle at any displacement $y$ is given by $F = m \omega^2 y$.
When the particle is midway between the mean and extreme positions, the displacement is $y = \frac{a}{2}$.
Substituting this value into the force equation:
$F = m \omega^2 \left(\frac{a}{2}\right) = \frac{1}{2} (m \omega^2 a)$.
Since $m \omega^2 a = 10 \,N$, we get:
$F = \frac{1}{2} \times 10 \,N = 5 \,N$.

(C) The acceleration of a particle executing $SHM$ is given by the formula:
$\alpha = -\omega^2 y$
where $\alpha$ is the acceleration,$\omega$ is the angular frequency,and $y$ is the displacement from the mean position.
From the relation $\alpha \propto y$,it is clear that the magnitude of acceleration is directly proportional to the displacement from the equilibrium position.
The maximum displacement of the particle from the mean position is equal to the amplitude $A$.
Therefore,the acceleration is maximum when the displacement $y$ is maximum (i.e.,at the extreme positions where $y = \pm A$).

(A) The displacement-time graph shown in the figure is a sine wave, so the equation of displacement is $x = A \sin(\omega t)$.
From the graph, the amplitude $A = 1 \,cm$ and the time period $T = 8 \,s$.
Therefore, the angular frequency $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{8} = \frac{\pi}{4} \,rad/s$.
The equation for displacement is $x = 1 \sin\left(\frac{\pi}{4} t\right)$.
The acceleration $a$ in simple harmonic motion is given by $a = -\omega^2 x = -\omega^2 A \sin(\omega t)$.
Substituting the values at $t = \frac{4}{3} \,s$:
$a = -\left(\frac{\pi}{4}\right)^2 \times 1 \times \sin\left(\frac{\pi}{4} \times \frac{4}{3}\right)$
$a = -\frac{\pi^2}{16} \times \sin\left(\frac{\pi}{3}\right)$
$a = -\frac{\pi^2}{16} \times \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{32} \pi^2 \,cm \,s^{-2}$.
Thus, the correct option is $A$.

(C) The magnitude of acceleration $a$ for a simple harmonic oscillator at a displacement $x$ from the mean position is given by the formula $a = \omega^2 x$.
Here,$\omega$ is the angular frequency,defined as $\omega = \frac{2 \pi}{T}$.
Given: Time period $T = 2 \ s$,displacement $x = 0.25 \ m$.
First,calculate the angular frequency: $\omega = \frac{2 \pi}{2} = \pi \ rad/s$.
Now,substitute the values into the acceleration formula: $a = (\pi)^2 \times 0.25$.
Since $0.25 = \frac{1}{4}$,we get $a = \pi^2 \times \frac{1}{4} = \frac{\pi^2}{4} \ m \ s^{-2}$.

(B) The acceleration of a particle in $SHM$ is given by $a(t) = -\omega^2 x = -\omega^2 A \cos(\omega t)$.
At the extreme position,$x = A$ $(t = 0)$,and at the equilibrium position,$x = 0$ $(t = \frac{\pi}{2\omega})$.
The average acceleration $A_{\text{avg}}$ is defined as $\frac{1}{\Delta t} \int_{0}^{\Delta t} a(t) dt$.
Here,$\Delta t = \frac{\pi}{2\omega}$.
$A_{\text{avg}} = \frac{1}{\pi / 2\omega} \int_{0}^{\pi / 2\omega} \omega^2 A \cos(\omega t) dt = \frac{2\omega}{\pi} [A \sin(\omega t)]_{0}^{\pi / 2\omega} = \frac{2\omega A}{\pi} (1 - 0) = \frac{2\omega^2 A}{\pi}$.
Since the maximum acceleration $A_{\max} = \omega^2 A$,the ratio is $\frac{A_{\text{avg}}}{A_{\max}} = \frac{2\omega^2 A / \pi}{\omega^2 A} = \frac{2}{\pi}$.