Question

The vertical motion of a ship at sea is described by the equation $\frac{d^{2}x}{d{t}^2}=-4x$ , where $x$ is the vertical height of the ship (in $\text{metre}$) above its mean position. If it oscillates through a maximum height of $1\text{m}$, then:

• A. Its maximum vertical speed will be $1{\text{ms}}^{-1}$
• B. Its maximum vertical speed will be $2{\text{ms}}^{-1}$
• C. Its greatest vertical acceleration is $2{\text{ms}}^{-2}$
• D. its greatest vertical acceleration is $1{\text{ms}}^{-2}$

Answer 1

The correct option is B Its maximum vertical speed will be $2{\text{ms}}^{-1}$

Given, equation of motion of ship
$\frac{d^{2}x}{d{t}^2}=-4x....\left(i\right)$
Where $x=$displacement of ship from its mean position.
$\Rightarrow a\propto -x$, hence it represents the simple harmonic motion.

Amplitude of oscillation of ship, $A=1\text{m}$
Comparing Eq.$\left(i\right)$ with $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$
$\Rightarrow {\omega }^2=4$
$\Rightarrow \omega =2\text{rad/s}$
for a body executing SHM,
$v_{\text{max}}=A\omega =2\text{m/s}$
and maximum acceleration is,
$a_{\text{max}}={\omega }^{2}A=4{\text{m/s}}^2$