Homework Help: physics

Posted by Nicole on Wednesday, March 7, 2012 at 11:47pm.

An object with mass 3.0 kg is attached to a spring with spring stiffness constant k = 300 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.
(a) Calculate the amplitude of the motion.

m
(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]
m/s

physics - drwls, Thursday, March 8, 2012 at 5:22am
Total energy (which is conserved)
= (1/2)M*V^2 + (1/2)k X^2
= 1.5*0.55^2 + 150*0.02^2 = 0.5138 J

(a) when V = 0 you get the maximum deflection Xmax. That is the amplitude.
0.5138 = (300/2)*Xmax^2
Xmax = 0.0585 m

(b) when X = 0 you have maximum speed:
0.5138 = 1.5*Vmax^2
Vmax = 0.5853 m/s

Answer this Question

Related Questions

physics - a 0.1 kg mass is suspended at rest from a spring near the Earth's...
physics - If a 0.1 kg mass is suspended at rest from a spring near the Earth&#...
PHYSICS - a 0.1 kg mass is suspended at rest from a spring near the Earth's ...
Physics - A vertical spring (ignore its mass), whose spring stiffness constant ...
Physics - An object with mass 3.7 is attached to a spring with spring stiffness ...
PHYSICS - A spring is hung from a ceiling, and an object attached to its lower ...
PHYSICS - A 0.49 kg object is attached to a spring with a spring constant 167 N/...
Physics - A 2.5 m rod of mass M= 4. kg is attached to a pivot .8m from the left ...
Physics - An ideal spring with a stiffness of 329 N/m is attached to a wall, and...
physics - a 0.75 kg mass attached to avertical spring streches the spring 0.30m...

For Further Reading

First Name:
School Subject:
Answer: