A) When a .20 kg mass is suspended from a vertically hanging spring it stretches

Question

A) When a .20 kg mass is suspended from a vertically hanging spring it stretches by .0098m. What is the constant of the spring?

B) the same mass attached to the same spring is now placed on a horizontal frictionless surface. The mass is then pulled from it's mean position so as to stretch the spring by 0.01m and is released at time t=0.
Calculate: Time period of subsequent motion, and write an expression that gives the correct displacement (0.01m at t=0) of the mass as a function of time.

C). Use the expression you found in part B to find the displacement of the mass 1/6 of a period after its released.


Is this right???
A) F=-kx
-k=mg/x
-k= (.20kg)(-9.8m/s2)/ .0098m

k= 200.2

T= 0.1986 sec

w=2/T

w= 2/.1986

w=31.6 rad/s

x=Acos(wt)

x=(0.01m)cos(31.6*0)

x= .01m

w= 2/(.167*.1986s)
w= 289.8

x=Acos(wt)

x=(0.01m)cos(31.6*0)

x=(.01m)cos(289.8*0)

x=.01m

Explanation / Answer

correct

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