Question

In the system as shown, mass of the block is 1kg and spring constant is 25 $\frac{N}{m}$. The maximum kinetic energy of the block is 50J.

• A. The amplitude of oscillation is 2m
• B. At half of the amplitude, the kinetic energy is 37.5 J
• C. At half of the amplitude, potential energy of the spring is 12.5 J
• D. At half of the amplitude, potential energy of the spring is 20 J

Answer 1

Answer: (A), (B), (C)

At half of the amplitude, potential energy of the spring is 12.5 J

For a system with only conservative forces present, the total mechanical energy is reserved. For a spring mass system, the kinetic energy K.E. is maximum at the mean position, where the potential energy P.E is zero. Similarly, at the extremes, P.E is maximum and K.E is zero. Conserving total M.E. = (K.E + P.E) at the mean and not same positions. We see-.

$M.E{.}_{mean}=M.E_{extreme}$

$\Rightarrow \frac{1}{2}mv(x-0{)}^2+0=0+\frac{1}{2}k{A}^2$, where x=a is the amplitude

$\Rightarrow \frac{1}{2}k{A}^2=\frac{1}{2}mv(x-0{)}^2$=maximum Kinetic energy

$\frac{1}{2}x25x{A}^2=50$ j(as given)

$\Rightarrow =\sqrt{(50x\frac{2}{25}m}=2m$ ...(1)

It is clear that the total M.E. = 50J, and will stay the same for every position x

From conservation of energy, K.E. = M.E - P.E = (50 - 12.5)J

= 37.5 J ...(2)

From our calculation, we see that options (a), (b) & (c) are correct