Question

A particle executing S.H.M. has an amplitude \(\mathrm{A}\) and periodic time \(\mathrm{T}\). The minimum time required by the particle to get displaced by \((\mathrm{A} / \sqrt{2})\) from its equilibrium position is \(\ldots \ldots \ldots \mathrm{s}\). (A) \(\mathrm{T}\) (B) \(\mathrm{T} / 4\) (C) \(\mathrm{T} / 8\) (D) \(\mathrm{T} / 16\)

Short answer

The minimum time required by the particle to get displaced by A/√2 from its equilibrium position is \(t = \frac{T}{8}\).

Step-by-step solution

Write the equation of SHM

A particle executing SHM can be represented by the equation: \(x(t) = A\sin(\omega t)\), where x(t) is the displacement from the equilibrium position at time t, A is the amplitude, and ω is the angular frequency. For a particle with a periodic time T, the angular frequency ω can be calculated as: \(\omega = \frac{2\pi}{T}\).

Write the displacement equation for the given displacement

Now we want to find the time when the displacement x(t) reaches A/√2. So we can set up the equation: \(\frac{A}{\sqrt{2}} = A\sin(\omega t)\).

Solve for t

To solve for t, we can first divide both sides of the equation by A and then take the inverse sine of both sides: \[\frac{1}{\sqrt{2}} = \sin(\omega t) \implies \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \omega t\] Now substitute the expression for ω and solve for t: \[\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{2\pi}{T}t \implies t = \frac{T}{2\pi}\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\] Use the fact that \(\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\), then solve for t: \[t = \frac{T}{2\pi}\times\frac{\pi}{4} = \frac{T}{8}\]

Choose the correct answer

Now we can see that the minimum time required by the particle to get displaced by A/√2 from its equilibrium position is T/8. Therefore, the correct answer is: (C) T/8

Key concepts

Amplitude

The concept of amplitude is essential in understanding simple harmonic motion (SHM). Imagine a pendulum swinging back and forth — its maximum distance from the center point represents the amplitude. In more technical terms, amplitude is the peak value of displacement from the equilibrium position. It tells us the extent of movement on either side of this rest position.

Key Features of Amplitude:

• It is a constant for a specific SHM and doesn't change unless external forces are applied.
• Amplitude determines the energy in the system; greater amplitude means more energy.
• In the context of the exercise, when the displacement is given as \( A/\sqrt{2} \), it represents a fractional amplitude, signifying a point that is less than the maximum but still significant in its motion.

Understanding the amplitude gives insight into the SHM's scale and energy levels and helps in solving problems like the one presented above.

Periodic Time

Periodic time, often referred to as the period, is a pivotal characteristic of SHM. It's the duration it takes for the particle to complete a full cycle of motion, returning to its starting point. Just as the Earth revolves around the Sun in one year, an SHM motion completes one full swing in its periodic time.

Characteristics of Periodic Time:

• Represented by the symbol \( T \), it is a crucial factor in determining how fast or slow an oscillating system is.
• The periodic time remains constant for a system unless the physical characteristics of the system change, such as mass or length in the case of pendulums.
• In the problem, knowing \( T \) helps us calculate how long it takes for the displacement to reach a given fraction of the amplitude, such as \( T/8 \) representing one-eighth of the full cycle.

Having a good grasp of periodic time is essential for predicting and understanding the behavior of any system that exhibits SHM.

Angular Frequency

Angular frequency connects the concepts of time and oscillation in SHM. It tells us how quickly oscillations happen, measured in radians per second. Imagine a spinning wheel — angular frequency describes how fast it spins around its center.

Features of Angular Frequency:

• It's denoted by \( \omega \) and relates directly to the periodic time through the formula \( \omega = \frac{2\pi}{T} \).
• An important trait is that higher angular frequency means the system oscillates more rapidly.
• In the exercise, understanding \( \omega \) enables us to find when the particle's displacement matches a particular value, such as \( A/\sqrt{2} \).

By determining angular frequency, we gain insight into the speed and efficiency of oscillations, crucial for solving SHM problems efficiently.