Q.Give examples of a one-dimensional motion where

The key idea here is understanding the nature of velocity in periodic motion, specifically whether the particle's direction of motion reverses or not. Simple Harmonic Motion (SHM) is a classic example of periodic motion involving velocity reversal.

(a) For a particle moving along the positive $x$-direction that comes to rest periodically and moves forward:

1. The particle's velocity must always be non-negative ($v(t) \ge 0$).
2. The velocity must periodically become zero.
3. An example is a particle whose velocity is described by $v(t) = v_0(1 - \cos(\omega t))$, where $v_0$ and $\omega$ are positive constants.
4. Here, $v(t) \ge 0$ always, and $v(t)=0$ periodically when $\cos(\omega t)=1$ (i.e., $\omega t = 2n\pi$). After momentarily coming to rest, $v(t)$ becomes positive again, so the particle moves forward.

(b) For a particle moving along the positive $x$-direction that comes to rest periodically and moves backward:

1. The particle's velocity must change sign periodically.
2. It moves in the positive $x$-direction ($v>0$), comes to rest ($v=0$), and then moves backward ($v<0$).
3. This behavior is characteristic of Simple Harmonic Motion (SHM). …

In one-dimensional motion, the direction of a particle's movement is determined by the sign of its velocity. If the velocity $v(t)$ is positive, the particle moves in the positive x-direction. If $v(t)$ is negative, it moves in the negative x-direction. When $v(t) = 0$, the particle is momentarily at rest.

Let's consider each case:

(a) The particle moving along positive x-direction comes to rest periodically and moves forward.

For a particle to always move forward (or be at rest), its velocity must always be non-negative, i.e., $v(t) \ge 0$. For it to come to rest periodically, $v(t)$ must become zero at regular intervals.

1. Choosing a velocity function:

   Consider the velocity function $v(t) = v_0(1 + \sin(\omega t))$, where $v_0$ and $\omega$ are positive constants.

   • Initial motion: At $t=0$, $v(0) = v_0(1 + \sin(0)) = v_0(1 + 0) = v_0$. Since $v_0 > 0$, the particle starts moving in the positive x-direction.
   • Periodically at rest: The particle comes to rest when $v(t) = 0$. This occurs when $1 + \sin(\omega t) = 0$, which means $\sin(\omega t) = -1$. This condition is met when $\omega t = \frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, \dots$. In general, $\omega t = \left(2n + \frac{3}{2}\right)\pi$ for $n = 0, 1, 2, \dots$. These are periodic times when the particle momentarily stops.
   • Moving forward: Since the sine function ranges from $-1$ to $1$, the term $1 + \sin(\omega t)$ ranges from $1 + (-1) = 0$ to $1 + 1 = 2$. Therefore, $v(t) = v_0(1 + \sin(\omega t))$ is always $\ge 0$. This means the particle never moves backward; it either moves forward or is momentarily at rest.

2. Position function (for completeness):

   If we assume the particle starts at $x=0$ at $t=0$, its position $x(t)$ can be found by integrating the velocity:

This position function describes a particle that continuously advances, with its speed periodically dropping to zero before increasing again, always maintaining a positive or zero velocity.

(b) The particle moving along positive x-direction comes to rest periodically and moves backward.

For a particle to move along the positive x-direction, come to rest, and then move backward, its velocity must start positive, become zero, and then become negative. This pattern must repeat periodically. This type of motion is characteristic of oscillations, such as Simple Harmonic Motion (SHM).

1. Choosing a position function:

   Consider a particle undergoing Simple Harmonic Motion described by the position function $x(t) = A \sin(\omega t)$, where $A$ is the amplitude ($A > 0$) and $\omega$ is the angular frequency ($\omega > 0$).

2. Deriving the velocity function:

   The velocity of the particle is the time derivative of its position:

3. Analyzing the motion: …

Concept: Physically-Motivated (Non-Sinusoidal) Constructions for Periodic-Rest Motion

Method: A Real Piecewise Scenario for (a); a Restoring-Force Argument for (b) — Neither Needs a Trigonometric Formula

Both stored answers reach for an explicit sinusoidal velocity/position function. This method instead builds (a) from a genuinely piecewise, physically-motivated scenario (no sine or cosine anywhere), and answers (b) with a qualitative force argument that shows why SHM-type reversal must happen for any linear restoring force, without ever writing down $x=A\sin\omega t$.

(a) Comes to rest periodically, always resumes forward

Physical scenario: a car in traffic, stopping fully at every red light, then accelerating forward again — never reversing.

1. Between consecutive lights (a distance apart, spaced evenly since the lights are evenly timed), the car's velocity profile is: starts at $v=0$ (light just turned green), rises smoothly to some cruising value, then decreases smoothly back to $v=0$ exactly at the next light (it comes to a full, genuine stop each time — not a rolling slowdown).

2. This can be built explicitly and piecewise, with no trigonometric function at all: over each block of duration $T$, take

a simple parabola in the local time $\tau$ within each block, zero at $\tau=0$ and $\tau=T$ (the light), and maximum $v_{\max}$ at the block's midpoint $\tau=T/2$. (This is exactly the same shape as the "monkey climbing a pole" velocity profile used elsewhere in this chapter — a repeating parabolic arch, not a sine wave.)

3. Check the two required conditions directly from this construction: $v(\tau)\ge0$ for every $\tau\in[0,T]$ (a downward parabola between two zeros is non-negative on that whole interval) — the car never reverses; and $v=0$ exactly at $\tau=0,T,2T,\ldots$ — periodic stops, at every light, by construction.

(b) Comes to rest periodically, then reverses — a restoring-force argument, no formula needed

Physical scenario: a block on a horizontal spring, pulled to one side and released.

4. State the one physical fact that drives everything: the restoring force always points back toward equilibrium, $F=-kx$ — whatever the sign of $x$, the force's sign is the opposite.

5. Consider the block released from $x=+A$ (displaced in the positive direction). The force is $F=-kA<0$ throughout the outward-then-return motion on the positive side — it can never push the block further into $+x$ once released; it only ever decelerates outward motion or accelerates inward motion.

6. Use energy, not calculus, to show the block MUST turn around. Total mechanical energy $E=\tfrac12kA^2$ is fixed (all potential at release, since $v=0$ there). As the block moves, if it ever reached some $x'>A$, that would require potential energy $\tfrac12kx'^2 > \tfrac12kA^2 = E$ — impossible, since kinetic energy cannot be negative. So the block can never travel past $x=A$ at all — meaning, released from rest at $x=A$ moving inward (toward the origin) is the only possible history consistent with energy conservation; by the same logic applied to the return swing, it cannot pass $x=-A$ on the far side either. It must decelerate to rest at some extremum, then reverse — this is forced by energy conservation before ever solving the equation of motion. …