Question 7: Oscillations

Question

A body is executing SHM. When its displacements from the mean position are $$4 \mathrm{~cm}$$ and $$5 \mathrm{~cm}$$ it has velocity $$10 \mathrm{~cms}^{-1}$$ and $$8 \mathrm{~cms}^{-1}$$ respectively. Its periodic time $$\mathrm{t}$$ is

Accepted Answer

When a body is executing Simple Harmonic Motion (SHM), its velocity $$v$$ at a displacement $$x$$ from the mean position can be given by the equation:
$$v = \omega \sqrt{A^2 - x^2}$$
Here, $$\omega$$ is the angular frequency and $$A$$ is the amplitude of the SHM.
Given:
At displacement $$4 \mathrm{~cm}$$, velocity $$10 \mathrm{~cms}^{-1}$$
At displacement $$5 \mathrm{~cm}$$, velocity $$8 \mathrm{~cms}^{-1}$$
We use the equation for velocity at displacement for both given conditions to form two equations:
At $$x = 4 \mathrm{~cm}$$ and $$v = 10 \mathrm{~cms}^{-1}$$:
$$10 = \omega \sqrt{A^2 - 4^2}$$
This can be rewritten as:
$$10 = \omega \sqrt{A^2 - 16}$$
At $$x = 5 \mathrm{~cm}$$ and $$v = 8 \mathrm{~cms}^{-1}$$:
$$8 = \omega \sqrt{A^2 - 5^2}$$
This can be rewritten as:
$$8 = \omega \sqrt{A^2 - 25}$$
Now, we square both equations and equate them to form a system of equations:
For the first equation:
$$100 = \omega^2 (A^2 - 16)$$
For the second equation:
$$64 = \omega^2 (A^2 - 25)$$
Divide the first equation by the second equation to eliminate $$\omega^2$$:
$$\frac{100}{64} = \frac{A^2 - 16}{A^2 - 25}$$
Simplify:
$$\frac{25}{16} = \frac{A^2 - 16}{A^2 - 25}$$
Cross-multiplying gives:
$$25(A^2 - 25) = 16(A^2 - 16)$$
Expand and simplify:
$$25A^2 - 625 = 16A^2 - 256$$
Rearrange terms:
$$9A^2 = 369$$
Therefore:
$$A^2 = 41$$
The amplitude $$A = \sqrt{41}$$.
We can now use either set of velocity and displacement to find $$\omega$$. Using the first set:
$$10 = \omega \sqrt{41 - 16}$$
$$10 = \omega \sqrt{25}$$
Therefore:
$$10 = 5\omega$$
$$\omega = 2 \mathrm{~rad/s}$$
The periodic time $$T$$ is given by the relation:
$$T = \frac{2\pi}{\omega}$$
Substitute $$\omega$$:
$$T = \frac{2\pi}{2} = \pi \mathrm{~sec}$$
Therefore, the periodic time $$\mathrm{t}$$ is
Option D: $$\pi$$ sec
.

Answer

$$\frac{2 \pi}{3} \sec$$

Answer

2$$\pi$$ sec

Answer

$$\frac{3 \pi}{2} \mathrm{sec}$$

Answer

$$\pi$$ sec

Oscillations

A body is executing SHM. When its displacements from the mean position are $4 \mathrm{~cm} and 5 \mathrm{~cm} it has velocity 10 \mathrm{~cms}^{-1} and 8 \mathrm{~cms}^{-1} respectively. Its periodic time \mathrm{t}$ is

Choose the correct answer: