COMEDK 2024 Shift 1 Paper

In simple harmonic motion (SHM), the total mechanical energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE). At any point in the motion, the kinetic energy and potential energy can be expressed as:
The total energy (E) in SHM of amplitude A is
E=‌

1

2

kA2
where k is the spring constant.
The potential energy (PE) at a distance x from the mean position is given by:
PE=‌

1

2

kx2
The kinetic energy (KE) at a distance x from the mean position is:
KE=‌

1

2

k(A2−x2)
We need to find the distance from the mean position where the kinetic energy is equal to the potential energy. Therefore, we set the kinetic energy equal to the potential energy:
‌

1

2

kx2=‌

1

2

k(A2−x2)
By simplifying the equation, we get:
x2=A2−x2
Adding x2 to both sides, we have:
2x2=A2
Dividing both sides by 2 , we get:
x2=‌

A2

2

Taking the square root of both sides, we find:
x=‌

A

√2

=‌

A

√2

×‌

√2

√2

=‌

A√2

2

=‌

√2

2

A
We know that √2≈1.41, so:
x≈‌

1.41

2

A≈0.71A
Therefore, the distance from the mean position at which the kinetic energy is equal to its potential energy is approximately 0.71A. Hence, the correct answer is Option B .