A particle executes simple harmonic motion with an amplitude of 4 cm. At what displacement from the mean position is its kinetic energy equal to its potential energy?

B. 2 2 cm — The kinetic energy is given by K = 12k(A^2 - x^2) and potential energy by U = 12kx^2. Setting K = U leads to A^2 - x^2 = x^2, which simplifies to x = A 2. Substituting A = 4 gives x = 4 2 = 2 2 cm.

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$A particle executes simple harmonic motion with an amplitude of $A = 4 \text{ cm}$. At what displacement $x$ from the mean position is its kinetic energy equal to its potential energy?$

A. $2 \text{ cm}$
B. $2\sqrt{2} \text{ cm}$
C. $\frac{4}{\sqrt{2}} \text{ cm}$
D. $1 \text{ cm}$

Show correct answer & step-by-step solution

Correct answer: B — $2\sqrt{2} \text{ cm}$

Solution

The kinetic energy is given by

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

and potential energy by

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

. Setting

K = U

leads to

A^{2} - x^{2} = x^{2}

, which simplifies to

x = \frac{A}{2}

. Substituting

A = 4 cm

gives

x = \frac{4}{2} = 22 cm

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