Question

A body is executing simple harmonic motion. When the displacements from the mean position is $4 \,cm$ and $5 \,cm$, the corresponding velocities of the body is $10 \,cm/sec$ and $8\, cm/sec$. Then the time period of the body is

• A. $\frac{2\pi}{2}$ sec
• B. $ \pi / 2 $ sec
• C. $ \pi $ sec
• D. $ ( 3 \pi / 2) $ sec

Answer 1

The Correct Option is C

To find the time period of the body executing simple harmonic motion (SHM), we use the properties of SHM where the displacement, velocity, amplitude, and angular frequency are involved.

The velocity v in SHM is given by:

v = \omega \sqrt{A^2 - x^2}

• \omega is the angular frequency.
• A is the amplitude of motion.
• x is the displacement from the mean position.

Given that:

• x_1 = 4\, \text{cm}, v_1 = 10\, \text{cm/s}
• x_2 = 5\, \text{cm}, v_2 = 8\, \text{cm/s}

Applying the velocity formula for each case:

For x_1 and v_1:

10 = \omega \sqrt{A^2 - 4^2}

For x_2 and v_2:

8 = \omega \sqrt{A^2 - 5^2}

Squaring both equations, we get:

100 = \omega^2 (A^2 - 16) → (1)

64 = \omega^2 (A^2 - 25) → (2)

Now, subtract equation (2) from equation (1):

100 - 64 = \omega^2 [(A^2 - 16) - (A^2 - 25)]

36 = \omega^2 (9)

Solving for \omega^2:

\omega^2 = 4

\omega = 2

The angular frequency \omega is related to the time period T by the equation:

\omega = \frac{2\pi}{T}

Substituting the known value of \omega:

2 = \frac{2\pi}{T}

Thus, T = \pi seconds.

Therefore, the correct answer is \pi sec.