Q 2) A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

Consider a simple haemonic motion of time period T . Calculate the time taken for the displacement to change value from half the amplitude to the amplitude.

KE is half of the total energy when the potential energy is also half the total. Since PE is proportional to displacement^2, the displacement would be sqrt(1/2) = 70.7% of the maximum value.

Please post followup questions separately

To find the distance from the mean position where the kinetic and potential energies are equal in simple harmonic motion, we need to first understand the relationship between kinetic energy (K.E.) and potential energy (P.E.) in simple harmonic motion.

In simple harmonic motion, the total mechanical energy of the system remains constant. Therefore, the sum of the kinetic energy and potential energy at any point in the motion is equal to the total mechanical energy.

Mathematically, this can be expressed as:
K.E. + P.E. = Constant

Since the total mechanical energy is constant, we can set the kinetic energy equal to the potential energy and solve for the distance from the mean position.

Let's denote the distance from the mean position as 'x'.

The potential energy of a particle undergoing simple harmonic motion is given by:
P.E. = (1/2) * k * x^2

where k is the force constant of the system.

The kinetic energy of the particle is given by:
K.E. = (1/2) * m * v^2

where m is the mass of the particle and v is its velocity.

Since the particle is at its maximum displacement (amplitude) from the mean position, the velocity is zero at this point. Therefore, the kinetic energy is also zero.

Setting the kinetic energy equal to the potential energy and solving for x:
0.5 * m * 0^2 = 0.5 * k * x^2

Simplifying the equation:
0 = k * x^2

Since k is a positive constant, the only way for this equation to be true is if x = 0. Therefore, at the mean position, where x = 0, the kinetic and potential energies are equal.

In conclusion, the distance from the mean position where the kinetic and potential energies are equal in simple harmonic motion is zero, i.e., at the mean position itself.