Question

A spring of negligible mass having a force constant k extends by an amount y when a mass m is hung from it. The mass is pulled down a little and released. The system begins to execute simple harmonic motion of amplitude A and angular frequency $\omega$. The total energy of the mass - spring system will be

• A. $\frac{1}{2}m{A}^2{\omega }^2$
• B. $\frac{1}{2}m{A}^2{\omega }^2+\frac{1}{2}k{y}^2$
• C. $\frac{1}{2}k{y}^2$
• D. $\frac{1}{2}m{A}^2{\omega }^2-\frac{1}{2}k{y}^2$

Answer 1

The correct option is B $\frac{1}{2}m{A}^2{\omega }^2+\frac{1}{2}k{y}^2$

Let L be the relaxed length of the spring and y the extension produced in it due to force mg so that ky = mg (i)
The displacement of the mass during oscillation is given by
$x=Asin(\omega t+\varphi )$
At the instant when the displacement is x
​KE of mass = $\frac{1}{2}m{V}^2=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2=\frac{1}{2}m{A}^2{\omega }^{2}co{s}^2(\omega t+\varphi )$ (iii)
PE of spring = $\frac{1}{2}k(y+x{)}^2=\frac{1}{2}k(y^2+2yx+x^2)$ $=\frac{1}{2}k{y}^2+kyx+\frac{1}{2}k{x}^2$
Using (i) and (ii) and $\omega =\sqrt{\frac{k}{m}}$, we have
PE of spring =$\frac{1}{2}k{y}^2+mgx+\frac{1}{2}m{\omega }^2{A}^{2}si{n}^2(\omega t+\varphi )$ (iv)
taking gravitational PE at the mean position to be zero.
Gravitational PE at x = -mg x (v)
Adding (iii), (iv) and (v), we get
Total energy of mass -spring system
=$\frac{1}{2}m{A}^2{\omega }^{2}co{s}^2(\omega t+\varphi )+\frac{1}{2}k{y}^2+mgx+\frac{1}{2}m{\omega }^2{A}^{2}si{n}^2(\omega t+\varphi )-mgx=\frac{1}{2}m{A}^2{\omega }^2+\frac{1}{2}k{y}^2$