a 5.00 kg partridge is suspended from a pear tree by an ideal spring of negligib

Question

a 5.00 kg partridge is suspended from a pear tree by an ideal spring of negligible mass. when the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20 s.

I found the speed as it passes through the equilibrium position to be 0.150 m/s

now I need to figure out:
what is the magnitude of the acceleration when it is 5.00*10^-2 m above the equilibrium position?

Explanation / Answer

T=2*pi*sqrt(m/k) 4.2=2*pi*sqrt(m/k) =>k=11.19N/m x=A cos(w*t) w=2*pi/T =1.4959ras/s v=-A*w sin(wt) a=-A*w^2cos(wt) A=0.1m when x= -5*10^-2 m t is t=0.7 sec a=-0.1*1.4959^2cos(1.4959*0.7)=0.11189 m/sec^2 upwards [i did some silly mistake previously see this solution]

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