Question

a particle is performing linear SHM at apoint A on its path its potential energy is three times its kinetic energy.At another point B on its path its kinetic enerrgy is thrree times its potential enenrgy.find the ratio of its potential energy at A to its potential enenrgy at B

Answer 1

Answer on Question #58414 - Physics – Mechanics

A particle is performing linear SHM at a point A on its path its potential energy is three times its kinetic energy. At another point B on its path its kinetic energy is three times its potential energy. Find the ratio of its potential energy at A to its potential energy at B.

Solution.

Let $T$ is a kinetic energy and $U$ is a potential energy. According to energy conservation law the whole of energy $W$ is constant and consists of kinetic and potential ones ($W = U + T$). So on the one hand:

$$U_A = 3T_A \quad \text{- at point A}$$ $$T_B = 3U_B \quad \text{- at point B}$$

And on another hand:

$$W = U_A + T_A$$ $$W = U_B + T_B$$ $$U_A + T_A = U_B + T_B$$

Let's use first equations

$$U_A + \frac{1}{3}U_A = U_B + 3U_B$$ $$\frac{4}{3}U_A = 4U_B$$ $$\frac{U_A}{U_B} = 3$$

Answer: 3.

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