Q.The motion of a particle of mass $m$ is given by $x = 0$ for $t < 0$ s, $x(t) = A \sin 4p\,t$ for $0 < t < (1/4)$ s ($A > o$), and $x = 0$ for $t > (1/4)$ s. Which of the following statements is true? (Note: more than one of the given options may be correct.)

Impulse-Momentum Theorem: $F=ma$ and impulse $=m\Delta v$.

$v(t)=4\pi A\cos 4\pi t$, $a(t) = -16\pi^2 A\sin 4\pi t$ for $0<t<1/4$ s.

(A) At $t=1/8$ s: $\sin(4\pi \cdot 1/8)=\sin(\pi/2)=1$, so $F=ma=-16\pi^2 Am$. True.

(B) At $t=0$: $v$ jumps from $0$ to $4\pi A$ (impulse $=m\cdot 4\pi A=4\pi Am$); at $t=1/4$ s: $v$ jumps from $-4\pi A$ to $0$ (impulse magnitude $=4\pi Am$ again). The option states the magnitude as $4\pi^2 Am$ — off by a factor of $\pi$ from the correct $4\pi Am$. False. …

The key to this problem is the Impulse-Momentum Theorem: the net impulse on a particle equals its change in momentum. When a particle starts from rest, moves, and then stops, impulses must act at the boundaries. Let's break it down.

1. Understanding the motion

   The position is given piecewise:

   • For $t < 0$: $x = 0$ (particle at rest at the origin).
   • For $0 < t < \frac{1}{4}$ s: $x(t) = A \sin(4\pi t)$.
   • For $t > \frac{1}{4}$ s: $x = 0$ (particle back at rest at the origin).

   The sine term has angular frequency $\omega = 4\pi$ rad/s. Over the interval $0$ to $\frac{1}{4}$ s, the argument $4\pi t$ goes from $0$ to $\pi$, so the particle executes exactly half a sine wave — starting at $x=0$, rising to $x=A$ at $t=1/8$ s, and returning to $x=0$ at $t=1/4$ s.

2. Velocity and acceleration

   Differentiate $x(t)$ to get velocity:

Differentiate again for acceleration:

At $t = 0^+$, $\sin(0)=0$ so $a=0$, but $v(0^+) = 4\pi A$ — the particle jumps from rest to a finite velocity instantly. That means an impulse acts at $t=0$. Similarly, at $t = \frac{1}{4}^-$, $\sin(\pi)=0$ so $a=0$, and $v(\frac{1}{4}^-) = 4\pi A \cos(\pi) = -4\pi A$. The particle then stops at $t = \frac{1}{4}^+$, so another impulse acts.

3. Checking option (A): Force at $t = 1/8$ s

   At $t = 1/8$ s, $\sin(4\pi \cdot \frac{1}{8}) = \sin(\pi/2) = 1$.

   So $a = -16\pi^2 A$.

   Force $F = ma = -16\pi^2 A m$.

   This matches option (A) exactly. True.

4. Checking option (B): Impulse magnitude at $t=0$ and $t=1/4$ s

   Impulse $J = \Delta p = m \Delta v$.

   At $t=0$: velocity jumps from $0$ to $4\pi A$, so $\Delta v = 4\pi A$.

   Impulse $J = m \cdot 4\pi A = 4\pi A m$. …

Concept: Impulse-momentum theorem applied to a discontinuous-velocity motion

Step 1: Get velocity and acceleration for $0<t<1/4$ s.

Step 2: Evaluate statement (a) — force at $t=1/8$ s.

$\sin(4\pi \cdot \tfrac18)=\sin(\pi/2)=1$, so $a=-16\pi^2A$ and

$F=ma=-16\pi^2Am$. This matches (a) exactly. (a) is TRUE.

Step 3: Evaluate statements (b) and (e) — the impulses at the endpoints.

Just before $t=0$, $v=0$; just after, $v(0^+)=4\pi A\cos 0=4\pi A$ — a

velocity jump, meaning an impulsive force acts at $t=0$. So (e) ("no

impulse") is FALSE. The magnitude of that impulse is

$J=m\Delta v = 4\pi A m$. Symmetrically, at $t=1/4$ s, $v$ jumps from

$-4\pi A$ to $0$, impulse magnitude $4\pi A m$ again. Option (b) claims the …