A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. The mass passes through the equilibrium point (y=0) with positive velocity at t=0. Positive direction of y-axis is downward.

Part A

What equation describes this motion as a function of time?

Part B

At what time will the spring stretch to its maximum length at first time?

Part C

At what time will the spring shrink to its minimum length at first time?

Question

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Respuesta :

boffeemadrid boffeemadrid · Answer 1 · 2020-01-27T07:41:34+01:00

Answer:

[tex]y(t)=0.28sin(36.02599t)[/tex]

0.0436 s

0.1308 s

Explanation:

A = Amplitude = 28 cm

m = Mass = 0.235 kg

k = Spring constant = 305 N/m

The equation which describes motion as a function of time is given by

[tex]y(t)=Asin(\omega t)[/tex]

Angular speed is given by

[tex]\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \omega=\sqrt{\dfrac{305}{0.235}}\\\Rightarrow \omega=36.02599\ rad/s[/tex]

The equation is

[tex]\mathbf{y(t)=0.28sin(36.02599t)}[/tex]

Maximum length will be at amplitude

Amplitude is given by

[tex]A=Asin(\omega t)[/tex]

here

[tex]\omega t=\dfrac{\pi}{2}\\\Rightarrow t=\dfrac{\pi}{2\times 36.02599}\\\Rightarrow t=0.0436\ s[/tex]

The time to stretch to maximum length is 0.0436 s

At minimum length

[tex]y(t)=-A\\\Rightarrow -A=Asin\omega t[/tex]

[tex]\omega t=\dfrac{3\pi}{2}\\\Rightarrow t=\dfrac{3\pi}{2\omega}\\\Rightarrow t=\dfrac{3\pi}{2\times 36.02599}\\\Rightarrow t=0.1308\ s[/tex]

The time will the spring shrink to its minimum length at first time is 0.1308 s