Physics: Avertical 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?

Yee Been there, didnt end up so wellExplanation:...

Avertical spring with spring stiffness constant

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16.01.2022

y(t)=0.28sin(36.02599t)

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

$y(t)=Asin(\omega t)$

Angular speed is given by

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

The equation is

$\mathbf{y(t)=0.28sin(36.02599t)}$

Maximum length will be at amplitude

Amplitude is given by

$A=Asin(\omega t)$

here

$\omega t=\dfrac{\pi}{2}\\\Rightarrow t=\dfrac{\pi}{2\times 36.02599}\\\Rightarrow t=0.0436\ s$

The time to stretch to maximum length is 0.0436 s

At minimum length

$y(t)=-A\\\Rightarrow -A=Asin\omega t$

$\omega t=\dfrac{3\pi}{2}\\\Rightarrow t=\dfrac{3\pi}{2\omega}\\\Rightarrow t=\dfrac{3\pi}{2\times 36.02599}\\\Rightarrow t=0.1308\ s$

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

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