A) obtain the equation of motion for a 125g mass at the end of a perfectly elast

Question

A) obtain the equation of motion for a 125g mass at the end of a perfectly elastic spring which when stretched 14.5 cm from equilibrium and then released from rest, undergoes a simple harmonic motion with a period of 0.8727s.

B) find the maximum velocity and total energy of the mass

PLEASE DON NOT GOOGLE AND THEN COPY AND PASTE THE YAHOO ANSWER. I ALREADY DID THIS AND LOOKED AT THE ANSWER HOWEVER I AM LOOKING FOR SOMEONE WHO CAN SIMPLIFY THIS AND GIVE ALL THE CORRECT ANSERS AND WORKINGS WITH THE PLUG AND CHUG COMPLETED.

THANKS IN ADVANCE

Explanation / Answer

time period = 2?*sqrt(m/k)

m= 125g

T = 0.8727 sec

w = 7.2 rad/s

spring constant k = 4?^2*m/T^2

= 4?^2*0.125/(0.8727^2)

= 6.5 N/m

eq of motion

x= 0.145*cos (7.2t) m

total energy = 0.5*k*Xmax^2

= 0.5*6.5*0.145^2

= 0.068 J

Vmax = Amax*w

Amax = max amplitude

Vmax = 0.145*7.2

= 1.044 m/s

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