- V = E, K = 0
Explanation:
Key concept: Energy Graph for a Spring: If the mass attached with spring performs simple harmonic motion about its mean position then its potential energy at any position (x) can be given by,
Now kinetic energy at any position $\text{K}=\text{E}-\text{U}=\frac{1}{2}\text{Ka}^2-\frac{1}{2}\text{Kx}^2$
$\text{K}=\frac{1}{2}(\text{a}^2-\text{x}^2)\ ....(\text{i})$
From the above formula, we can cheak that
$\text{U}_\text{max}=\frac{1}{2}\text{Ka}^2$ $\big[\text{At extreme x}=\pm\text{a}\big]$ and Umin = 0 [At mean x = 0]
$\text{K}_\text{max}=\frac{1}{2}\text{Ka}^2$ [At mean x = 0] and Kmin = 0 $\big[\text{At extreme x}=\pm\text{a}\big]$
$\text{E}=\frac{1}{2}\text{ka}^2=\text{constant(at all positions)}$
Because velocity of mass = 0 [at extreme position]
$\therefore\ \text{K}=\frac{1}{2}\text{mv}^2=0$
It means kinetic energy changes parabolically w.r.t. position but total energy remain always constant irrespective to the position of mass.
Total energy is E = PE + KE, which remains constant.
According to the question, when paraticle is at x = xm, i.e., at extreme position x = xm ⇒ KE = 0
Hence, total energy will be
$\text{E}=\text{PE}+0=\text{PE}$
$\Rightarrow\text{V}\text{(x}_\text{m})=\frac{1}{2}\text{kx}^2_\text{m}$
Hence option (b) is correct.
V = E, K = 0
