Question

1
2. In one complete oscillation, the P.E. and K.E. of a simple
harmonic oscillator attain maximum value
twice
thrice
none​

Answer 1

Answer:

twice mmmmmmmmmmookkjhcsagnbdsfh csdgjmn

Answer 2

Answer:

The correct answer is twice.

Explanation:

In Simple harmonic motion or SHM, the respective potential energy and kinetic energy of the simple harmonic oscillator is given by the following given formula:

P.E = $\frac{1}{2} mx^{2} y^{2}$

K.E = $\frac{1}{2} mx^{2} (a^{2} -y^{2})$

Here x = ω which is the angular frequency of the oscillator, m is the mass of bob attached to the respective spring, and "y" here is the displacement of the bob from the mean position. Also, "a" here is the maximum displacement of the bob during SHM.

Kinetic energy is maximum or greatest whenever y = 0 which is when the bob comes to the mean position. Therefore the kinetic energy attains the maximum greatest value twice during SHM.

The potential energy from the formula stated above, we can see that potential energy will be maximum at y = a that is at the extreme positions of the bob twice. Hence potential energy also attains the greatest maximum value twice.

So the answer among the 3 given options is twice.

The kinetic and potential energy in one complete oscillation attains the maximum value two times or twice.