Question

#22. Answer is A.
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A 2.0-kg block oscillates on the end of a mass less spring. The resulting simple harmonic motion is shown in the drawing what is the potential energy of the block at t = 1.0 s? 0.016J 4.0J 13J 0.72J n.o.t

Answer 1

Time Period from the graph is T= 4.0s.

Mass of the block, m= 2.0kg

Now if "k" is the spring constant then Time Period is given by

  $T= 2 \pi \sqrt{\frac{m}{k}}$

Using all the given values in above,

$4.0= 2 \times 3.14 \sqrt{\frac{2.0}{k}}$

$\sqrt{\frac{2.0}{k}}= 0.637$

Squaring both sides, we get

$\frac{2.0}{k}= 0.406$

or Spring constant, $k= \frac{2.0}{0.406}= 4.926 N/m$

Now at time t= 1.0s, displacement is x= 0.080m

then Potential Energy of the system at this time will be

$P.E= \frac{1}{2}kx^2= \frac{1}{2} \times 4.926 \times (0.08)^2$

  $P.E= 0.01576 J\sim 0.016J$(ANS)