A block of mass 5 kg moves along the x-direction subject to the force F = (−20x + 10) N, with the value of x in metre. At time t= 0 s, it is at rest at position x = 1 m. The position and momentum of the block at t = (pi/4) s are

 The Harmonic Dance of a Block: A Journey Through Force and Motion

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Question Revisited

A block of mass $5\text{kg}$ moves along the $x$-axis under the influence of the force $F=(-20x+10)\text{N}$, where $x$ is in meters. At $t=0\text{s}$, the block is at rest at $x=1\text{m}$. Determine its position and momentum at $t=\frac{\pi }{4}\text{s}$.
Options:
(A)
(B)
(C)
(D)

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Historical Context

The study of oscillatory motion dates back to Galileo’s observation of chandeliers swinging in Pisa Cathedral, leading to the concept of simple harmonic motion (SHM). The force law $F=-kx$ governs SHM, but here, the force $F=-20x+10$ introduces a twist—a shifted equilibrium point. This mirrors real-world systems like pendulums in accelerating frames or charged particles in electric fields with constant offsets.

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Understanding the Force

The force $F=-20x+10$ can be rewritten as:

$$F=-20(x-0.5),$$

revealing a shifted harmonic oscillator with:

• Equilibrium position: $x_{\text{eq}}=0.5\text{m}$,

• Effective spring constant: $k=20\text{N/m}$.

The block oscillates around $x=0.5\text{m}$, not $x=0$.

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Equation of Motion

For SHM, angular frequency $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{20}{5}}=2\text{rad/s}$.
Let $x^{\mathrm{\prime }}(t)=x(t)-x_{\text{eq}}$ (displacement from equilibrium). The solution is:

$$x^{\mathrm{\prime }}(t)=A\cos (\omega t+\varphi ),$$

where $A$ is the amplitude and $\varphi$ is the phase.

Initial Conditions:

• At $t=0$: $x=1\text{m}\Rightarrow x^{\mathrm{\prime }}=1-0.5=0.5\text{m}$,

• Velocity $v(0)=0\Rightarrow \frac{d{x}^{\mathrm{\prime }}}{dt}=-A\omega \sin (\varphi )=0\Rightarrow \varphi =0$.

Thus, $x^{\mathrm{\prime }}(t)=0.5\cos (2t)$, so:

$$x(t)=0.5+0.5\cos (2t)\mathrm{.}$$

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Position at $t=\frac{\pi }{4}\text{s}$

Substitute $t=\frac{\pi }{4}$:

$$x\left(\frac{\pi }{4}\right)=0.5+0.5\cos (2\cdot \frac{\pi }{4})=0.5+0.5\cos \left(\frac{\pi }{2}\right)\mathrm{.}$$

Since $\cos \left(\frac{\pi }{2}\right)=0$:

$$x\left(\frac{\pi }{4}\right)=0.5\text{m}\mathrm{.}$$

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Momentum at $t=\frac{\pi }{4}\text{s}$

Velocity $v(t)=\frac{dx}{dt}=-0.5\cdot 2\sin (2t)=-\sin (2t)$.
At $t=\frac{\pi }{4}$:

$$v\left(\frac{\pi }{4}\right)=-\sin (2\cdot \frac{\pi }{4})=-\sin \left(\frac{\pi }{2}\right)=-1\text{m/s}\mathrm{.}$$

Momentum .

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Real-World Analogy

Imagine a spring attached to a cart on a track, with the spring’s equilibrium shifted by a constant force (e.g., a fan blowing air). The cart oscillates around the new equilibrium, just like this block. The math mirrors quantum harmonic oscillators or AC circuits with DC offsets!

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Critical Analysis of Options

• (A) Incorrect: Negative position contradicts $\cos$-based oscillation.

• (B) Incorrect: Zero momentum implies zero velocity, but $\sin (\pi \mathrm{/}2)\mathrm{\ne }0$.

• (C) Correct: Matches $x=0.5\text{m}$ and .

• (D) Incorrect: Position $-1\text{m}$ violates equilibrium at $0.5\text{m}$.

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Final Answer:

$$$$

Correct Option: (C)