Question

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the equilibrium position. (a) Find the position x of the mass at the times t = π/12, π/8, π/6, π/4, and 9π/32 s. (Use g = 32 ft/s2 for the acceleration due to gravity.)

Answer 1

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as$x(t)=Asin(\omega t+\phi)$

'A' is the amplitude = 6 inches (given)

$\omega =\sqrt{\frac{k}{m}}$ is the natural frequency of the system

At equilibrium we have

$mg=kx\\\\k=\frac{mg}{x}$

Applying values we get

$k=40 lb/ft$

thus natural frequency equals

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Thus the equation of motion becomes

$x(t)=6sin(8t+\phi)$

At time t=0 since mass is at it's maximum position thus we have

$A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})$

Thus the position of mass at the given times is as follows

1) at $\frac{\pi}{12}$ $x(t)=5.99inches$

2) at $\frac{\pi}{8}$ $x(t)=5.9909inches$

3) at $\frac{\pi}{6}$ $x(t)=5.98397inches$

4) at $\frac{\pi}{4}$ $x(t)=5.9639inches$

5) at $\frac{9\pi}{32}$ $x(t)=5.954inches$

Answer 2

The position x of the mass at the times t = π/12, π/8, π/6, π/4, and 9π/32 s is mathematically given a

•  $\frac{\pi}{12} xt =5.99in$

•   $\frac{\pi}{8} xt=5.9909in$

•    $\frac{\pi}{6} xt=5.98397in$

•    $\frac{\pi}{4} xt=5.9639in$

•   $\frac{9\pi}{32} xt=5.954in$

What is the position x of the mass at the times t = π/12, π/8, π/6, π/4, and 9π/32 s

Question Parameter(s):

A mass weighing 20 pounds

stretches a spring 6 inches.

The mass is initially released from rest from a point of 6 inches

g = 32 ft/s2 for the acceleration due to gravity

Generally, the equation for the simple harmonic motion  is mathematically given as

$x(t)=Asin(\omega t+\phi)$

And natural frequency

$\omega =\sqrt{\frac{k}{m}}$

Therefore

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Where

$x(t)=6sin(8t+\phi)$

when the time is t = 0

$A=Asin(\omega t+\phi)\\\\\(\omega\times 0+\phi)=1$

p=π/2

Hence

x(t)=Asin(wt+π/2)

In conclusion, the position of mass at the given times are  

•  $\frac{\pi}{12} xt =5.99in$

•   $\frac{\pi}{8} xt=5.9909in$

•    $\frac{\pi}{6} xt=5.98397in$

•    $\frac{\pi}{4} xt=5.9639in$

•   $\frac{9\pi}{32} xt=5.954in$

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