Question

In figure (12-E3) $k=100N{m}^{-1},M=1kg$ and $F=10N.$ (a) Fine 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

Given, $k=100N/m,m=1kg$ and $F=10N$

(a) In the equilibrium position,

Compression = $=\delta =\frac{F}{k}=\frac{k}{100}$

$=0.1m=10cm$

(b) The below imparts a speed of 2 m/s to the block towards left.

$\therefore P.E.+K.E.=\frac{1}{2}k{\delta }^2+\frac{1}{2}M{v}^2$

$=\frac{1}{2}\times 100\times (0.1{)}^2+\frac{1}{2}\times 1\times 4$

$=0.5+2=2.5J$

(c) Time period $2\pi \sqrt{\frac{M}{k}}$

$=2\pi \sqrt{\frac{M}{100}}=\frac{\pi }{5}sec$

(d) Let the amplitude be 'x' which means the distance between the mean position and the extreme position.

So, in the extreme position, compression of the spring is $(x-K).$

Since in SHM the total energy remains constant.

$\frac{1}{2}k(x+\delta {)}^2=\frac{1}{2}k{\delta }^2+\frac{1}{2}m{v}^2+Fx$

$=2.5+10x$

[Because $\frac{1}{2}k{\delta }^2+\frac{1}{2}m{v}^2=2.5]$

So, $50(x+0.1{)}^2=2.5+10x$

$\therefore 50x^2+0.5+10x=2.5+10x$

$\therefore 50x^2=2$

$\Rightarrow x^2=\frac{2}{50}=\frac{4}{100}$

$\Rightarrow x=\frac{2}{100}m=20cm$

(e) Potential Energy at the left extreme is given by,

$P.E.=\frac{1}{2}K(X+\delta {)}^2$

$=\frac{1}{2}\times 100\times (0.1+0.2{)}^2$

$=50\times \left(0.09\right)=4.5J$

(f) Potential Energy at the right extreme is given by,

$P.E.=\frac{1}{2}k(x+\delta {)}^2-F(2x)$

$\{2x=\text{distance between two extremes}\}$

$=4.5-10\left(0.4\right)=0.5J$

The different values in (b), (e) and (f) do not violate law of conservation of energy as the work is done by the external force 10 N.