A particle is executing SHM at mid point of mean position and extremity. What is the potential energy in terms of total energy

Question

A particle is executing SHM at mid point of mean position and extremity. What is the potential energy in terms of total energy (A)  (E/4)  (B)  (E/16)  (C)  (E/2)  (D)  (E/8)

Tardigrade Admin · Accepted Answer

If a particle executes SHM, its kinetic energy is given by

K E = \frac{1}{2} m ω^{2} (A^{2} - x^{2})

Or

K E = \frac{1}{2} k (A^{2} - x^{2})

where,

k = m ω^{2} =

constant
Its potential energy is given by

PE = \frac{1}{2} m ω^{2} x^{2}

= \frac{1}{2} k x^{2}

Thus, total energy of particle

E = K E + PE

= \frac{1}{2} k (A^{2} - x^{2}) + \frac{1}{2} k x^{2}

= \frac{1}{2} k A^{2}

Hence,

PE = \frac{1}{2} k x^{2} = \frac{1}{2} k (\frac{A}{2})^{2}

(∵ x = A /2)

= \frac{1}{4} (\frac{1}{2} k A^{2})

= \frac{1}{4} E

Hence, potential energy is one-fourth of total energy.

Q. A particle is executing SHM at mid point of mean position and extremity. What is the potential energy in terms of total energy

$\frac{E}{4}$

$\frac{E}{16}$

$\frac{E}{2}$

$\frac{E}{8}$

Q. A particle is executing SHM at mid point of mean position and extremity. What is the potential energy in terms of total energy

4E​

16E​

2E​

8E​

$\frac{E}{4}$

$\frac{E}{16}$

$\frac{E}{2}$

$\frac{E}{8}$