When a 0.20-kg block is suspended from a vertically hanging spring, it stretches

Question

When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from
its original length of 0.050 m to 0.060 m. The same block is attached to the same spring and
placed on a horizontal, frictionless surface. The block is then pulled so that the spring stretches
to a total length of 0.10 m. The block is released at time t = 0 s and undergoes simple harmonic
motion.
A) What is the maximum acceleration of the block?
B) What is the Frequency of the motion in Hertz?

Explanation / Answer

kx=mg x=0.6=0.5=0.1 k= 0.2*9.8/0.1=19.6 Max force for 0.1m elongation= k*0.1=m*a A>>>We get, a=k*0.1/m= 9.8m/s^2 ?=v(k/m) = 14rad/s B>>>>frequency=?/2p = 2.23 Hz C>>> Total Mechanical Energy at any instant: 0.5*k*(amplitude)^2 =0.098J

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