a massless spring of spring constant 50 N/m is suspended vertically. Set the ver

Question

a massless spring of spring constant 50 N/m is suspended vertically. Set the vertical origin to the natural length of the spring. A 0.5 kg mass is now attached to the end of the spring and let go at t=0.
a. What is the new equilibrium position for the mass hanging from the string?
b. what are the amplitude and frequency of the oscillations?
c. What is the maximum speed for the mass?

Explanation / Answer

a> At equilibrium position Net Force=0 k*x0=mg x0=.098 below origin b>At extreme position, v=0 equating Spring energy and potential energy 0.5k*a^2= -0.5*k*a^2+m*g*(2*a) amplitude=a=.1m c>Equating Energy at extreme(only potential) with energy at equilibrium position(only kinetic) 0.5*k*a^2+m*g*a=(0.5*m*v^2) v=sqrt(3)=1.732m/s (approx.)

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