Question 336: Oscillations

Question

A particle excuting S.H.M starts from the mean position. Its amplitude is 'A' and time period '$$\mathrm{T}$$' At what displacement its speed is one-fourth of the maximum speed?

Accepted Answer

To determine the displacement at which the speed of the particle is one-fourth of its maximum speed, we need to understand the relationship between displacement, amplitude, and velocity in Simple Harmonic Motion (SHM).
In SHM, the velocity $v$ of a particle at a displacement $x$ from the mean position is given by:
$$ v = \omega \sqrt{A^2 - x^2} $$
where:
$ \omega $ is the angular frequency, and
$ A $ is the amplitude.
The angular frequency $ \omega $ is related to the time period $ T $ by:
$$ \omega = \frac{2\pi}{T} $$
The maximum speed $ v_{max} $ occurs when the particle is at the mean position (i.e., $ x = 0 $), and it is given by:
$$ v_{max} = \omega A $$
We are given that the speed is one-fourth of the maximum speed. Therefore, we can write:
$$ \frac{v_{max}}{4} = \frac{\omega A}{4} $$
We need to find the displacement $ x $ at which this condition holds:
$$ \frac{\omega A}{4} = \omega \sqrt{A^2 - x^2} $$
We can divide both sides by $ \omega $:
$$ \frac{A}{4} = \sqrt{A^2 - x^2} $$
Next, we square both sides to eliminate the square root:
$$ \left(\frac{A}{4}\right)^2 = A^2 - x^2 $$
This simplifies to:
$$ \frac{A^2}{16} = A^2 - x^2 $$
Next, we rearrange the equation to solve for $ x $:
$$ x^2 = A^2 - \frac{A^2}{16} $$
$$ x^2 = \frac{16A^2 - A^2}{16} $$
$$ x^2 = \frac{15A^2}{16} $$
$$ x = \frac{A \sqrt{15}}{4} $$
Therefore, the displacement at which the speed is one-fourth of the maximum speed is:
Option D:
$$ \frac{\mathrm{A} \sqrt{15}}{40} $$

Answer

$$\frac{\mathrm{A}}{\sqrt{15}}$$

Answer

$$\frac{\mathrm{A}}{4}$$

Answer

$$\frac{4 \mathrm{~A}}{15}$$

Answer

$$\frac{\mathrm{A} \sqrt{15}}{40}$$

Oscillations

A particle excuting S.H.M starts from the mean position. Its amplitude is 'A' and time period '$\mathrm{T}$' At what displacement its speed is one-fourth of the maximum speed?

Choose the correct answer: