A body is executing simple harmonic motion. When the displacements from the mean position are 4 cm and 5 cm, the corresponding velocities of the body are 10 cm/s and 8 cm/s. Then the time period of the body is:
Detailed Answer:
For a body executing simple harmonic motion, the velocity \(v\) at a displacement \(x\) is given by the equation:
\[
v = \sqrt{\omega^2 (A^2 - x^2)}
\]
where:
- \( \omega = \frac{2\pi}{T} \) is the angular frequency,
- \( A \) is the amplitude,
- \( T \) is the time period,
- \( x \) is the displacement,
- \( v \) is the velocity.
Using the given data:
- At \( x = 4 \, \text{cm}, v = 10 \, \text{cm/s} \),
- At \( x = 5 \, \text{cm}, v = 8 \, \text{cm/s} \).
Substituting into the velocity equation:
\[
10 = \sqrt{\omega^2 (A^2 - 4^2)}
\]
\[
8 = \sqrt{\omega^2 (A^2 - 5^2)}
\]
Squaring both equations:
\[
100 = \omega^2 (A^2 - 16)
\]
\[
64 = \omega^2 (A^2 - 25)
\]
Dividing the equations:
\[
\frac{100}{64} = \frac{A^2 - 16}{A^2 - 25}
\]
Simplifying:
\[
\frac{25}{16} = \frac{A^2 - 16}{A^2 - 25}
\]
Cross-multiplying:
\[
25 (A^2 - 25) = 16 (A^2 - 16)
\]
Expanding:
\[
25A^2 - 625 = 16A^2 - 256
\]
Simplifying:
\[
9A^2 = 369 \quad \implies \quad A^2 = 41
\]
Substituting \( A^2 = 41 \) into \( \omega^2 (A^2 - 16) = 100 \):
\[
\omega^2 (41 - 16) = 100 \quad \implies \quad \omega^2 = \frac{100}{25} = 4
\]
Therefore:
\[
\omega = 2 \quad \text{and} \quad T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \, \text{seconds}.
\]
The time period of the body is:
\[
\boxed{\pi \, \text{seconds}}.
\]
.png)
0 Comments