Saturday

March 28, 2015

March 28, 2015

Posted by **noah** on Tuesday, July 26, 2011 at 11:22pm.

(a)how long does it take to reach the equilibrium position again?

(b)Write an equation giving its position y as a function of time t.

(c)What will be its maximum speed and maximum acceleration?

- college physic SHM -
**drwls**, Wednesday, July 27, 2011 at 1:50amThe spring constant is

k = Mg/0.215m = 57.0 N/m

The frequency of oscillation is

f = [1/(2 pi)]sqrt(k/m) = 1.075 Hz

The angular frequency is

w = sqrt(k/m) = 6.75 rad/s

The period is 1/f = 0.93 seconds

(a) 1/4 period = 0.2325 s is the time needed to return to the equilibrium position

(b) If y is measured from the equilibrium position

y = 0.13 cos wt = 0.14 cos6.75t

(c) Max speed = w*(amplitude)= 0.13 w

Max acceleration = w^2*(amplitude)

**Answer this Question**

**Related Questions**

college physic - A 1.25kg mass stretches a vertical spring 0.215m.If the spring ...

Physics - .A mass of 1.61 kg stretches a vertical spring 0.313 m. If the spring ...

Physics - A mass of 1.66 kg stretches a vertical spring 0.312 m. If the spring ...

College Physics PLEASE HELP - A spring stretches 0.150 m when a 0.30 kg mass is ...

Physics - A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The ...

Physics 203 - A block with mass of 5.0kg is suspended from an ideal spring ...

physics - A mass of 0.52 kg attached to a vertical spring stretches the spring ...

physic - Two equal masses are attached to separate identical springs next to one...

Physics - A mass of 0.55 kg attached to a vertical spring stretches the spring ...

physics - One end of a massless coil spring is suspended from a rigid ceiling. ...