Question
A block is resting on a piston which is moving vertically with a simple harmonic motion of period 1sec. At what amplitude of motion will the block and the piston seperate? What is the maximum velocity of the piston at this amplitude?
✓
Answer
We know that $\text{y}=\text{a}\sin\omega\text{t}$$\therefore$ Velocity of the block $=\frac{\text{dy}}{\text{dt}}=\text{a}\omega\cos\omega\text{t}$
Acceleration of the block $=\frac{\text{d}^2\text{y}}{\text{dt}^2}=\omega^2\sin\omega\text{t}$
$=-\omega\text{y}$
For maximum acceleration y = a$\therefore\Big(\frac{\text{d}^2\text{y}}{\text{dt}^2}\Big)_\text{max}$
$=-\omega\text{a}$
The block will be separated form the piston when$\omega^2\text{a}=\text{g}$
$=\text{a}=\frac{\text{g}}{\omega}$
$=\Big(\therefore\frac{2\pi}{\text{T}}\Big)$
$=\text{a}=\frac{\text{gT}^2}{4\pi^2}$ According to the given problem T = 1sec.$\therefore\text{a}=\frac{\text{g}}{4\pi^2}$
$=\frac{9.8}{4\times(3.14)^2}$
$=0.248\text{m/ sec}^2$
At this amplitude, the maximum velocity of the block will be$\omega\text{a}=\frac{2\pi}{\text{T}}$
$\frac{2\times3.14\times.0.248}{1}$
$=1.56\text{m/ sec}$
Acceleration of the block $=\frac{\text{d}^2\text{y}}{\text{dt}^2}=\omega^2\sin\omega\text{t}$
$=-\omega\text{y}$
For maximum acceleration y = a$\therefore\Big(\frac{\text{d}^2\text{y}}{\text{dt}^2}\Big)_\text{max}$
$=-\omega\text{a}$
The block will be separated form the piston when$\omega^2\text{a}=\text{g}$
$=\text{a}=\frac{\text{g}}{\omega}$
$=\Big(\therefore\frac{2\pi}{\text{T}}\Big)$
$=\text{a}=\frac{\text{gT}^2}{4\pi^2}$ According to the given problem T = 1sec.$\therefore\text{a}=\frac{\text{g}}{4\pi^2}$
$=\frac{9.8}{4\times(3.14)^2}$
$=0.248\text{m/ sec}^2$
At this amplitude, the maximum velocity of the block will be$\omega\text{a}=\frac{2\pi}{\text{T}}$
$\frac{2\times3.14\times.0.248}{1}$
$=1.56\text{m/ sec}$
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