Question

In figure $k=100N{m}^{-1}$, $M=1kg$ and $F=10N$, (a) Find the compression of the spring in the equilibrium position. (b) A sharp blow by some external agent imparts a speed of $2m{s}^{-1}$ to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant (c) Find the time period of the resulting simple harmonic motion. (d) Find the amplitude. (e) Write the potential energy of the spring when the block is at the left extreme. (f) Write the potential energy of the spring when the block is at the right extreme.
The answers of (b), (e) and (f) are different. Explain why this does not violate the principle of conservation of energy.

Answer 1

(a) At Equilibrium,

$Kx-7=0$

$Kx=7$

$x=7/K=10/100=0.1m$

(b) $PE+KE=\frac{1}{2}m{v}^2+\frac{1}{2}K{x}^2$

$=\frac{1}{2}\left(1\right)(2{)}^2+\frac{1}{2}(100\left)\right(0.1{)}^2$

$=\frac{5}{2}J$

(c) we know that,

in SHM $T=2\pi \sqrt{\frac{m}{k}}$

$T=2\pi \sqrt{\frac{1}{100}}=\frac{\pi }{5}=0.628s$

(d) As we know that,

At equilibrium position,

total energy of SHM $=\frac{1}{2}m{v}^2$

At turning points,

Total energy of $SHM=\frac{1}{2}k(x{)}^2=\frac{1}{2}K{A}^2$

Equating the two , we get

$\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2\Rightarrow A=v\sqrt{\frac{m}{k}}=2\sqrt{\frac{1}{100}}$

$A=0.2m$

(e) At left extreme , $PE=\frac{1}{2}k(A+x{)}^2=2J+\frac{1}{2}J+2J=\frac{9}{2}J$

(f) At right extreme, $PE=\frac{1}{2}k(x-A{)}^2=\frac{1}{2}\times 100\times (0.1{)}^2=\frac{1}{2}J$

This does not violate the principle

of conservation of energy as

the elongation of string is increasing,

the external force is ding more work.