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Chapter 11: Problem 8

A particle perform simple harmonic motion. The equation of its motion is \(x=5 \sin \left(4 t-\frac{\pi}{6}\right)\). Where \(x\) is its displacement. If the displacement of the particle is 3 units then its velocity is \(\quad\) [MP PMT 1994] (a) \(2 \pi / 3\) (b) \(5 \pi / 6\) (c) 20 (d) 16

Short Answer

Expert verified
The velocity when the displacement of the particle is 3 units is 16 units. So, the correct answer is (d) 16.

Step by step solution

01

Identify Variables

From the simple harmonic motion equation \(x=5 \sin \left(4t-\frac{\pi}{6}\right)\), we identify that the amplitude \(a = 5\) and the angular frequency \(\omega = 4\). Thus, \(a = 5\) and \(\omega = 4\).
02

Insert Identified Variables into Velocity Formula

Next, we need to insert the identified values into the velocity formula \(v=\omega \sqrt{a^2 - x^2}\) where \(v\) is the velocity, \(a\) is the amplitude, and \(\omega\) is the angular frequency. We are given that \(x = 3\) units when we need to find the velocity. So, we substitute \(a = 5\), \(\omega = 4\), and \(x = 3\) into the formula and solve.
03

Calculate the Velocity

When the values are entered into the formula, we obtain the equation \(v = 4 \sqrt{5^2 - 3^2}\). This simplifies to \(v = 4 \sqrt{25 - 9}\), which further simplifies to \(v = 4 \sqrt{16}\). Solving this gives \(v = 4 * 4 = 16\) units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
In simple harmonic motion, the amplitude is a crucial element. It represents the maximum displacement of the particle from its equilibrium or mean position. In simpler words, it's the farthest point the particle reaches from the central rest point during oscillation.
  • The amplitude sets the boundary for the motion of the particle.
  • It helps in determining the total range or extent of the motion.
  • Amplitude is always a positive value, as it signifies distance.
In the given equation, \(x = 5 \sin \left(4t - \frac{\pi}{6}\right)\), the amplitude is clearly indicated as 5 units. This means the particle moves 5 units from the resting position either to the extreme left or right.
Angular Frequency
Angular frequency describes how quickly the particle oscillates in simple harmonic motion. It's fundamentally different from regular frequency because it considers the angle traversed in each oscillation cycle rather than just the number of cycles per unit time.
  • Angular frequency is often denoted by \(\omega\).
  • Measured in radians per second.
  • Calculates how fast the oscillation occurs.
The equation provided, \(x = 5 \sin \left(4t - \frac{\pi}{6}\right)\), states that the angular frequency \(\omega = 4\). This indicates the particle completes a full oscillation in less time, leading to faster motion compared to lower angular frequencies.
Velocity Formula
The velocity of a particle in simple harmonic motion can be determined through a specific formula that incorporates both the angular frequency and the amplitude. The formula is as follows: \[v = \omega \sqrt{a^2 - x^2}\]
  • \(v\) is the velocity, which indicates how fast the particle is moving at any instant.
  • \(\omega\) is the angular frequency, illustrating the rate of oscillation.
  • \(a\) is the amplitude, reflecting the maximum displacement.
  • \(x\) is the current displacement at which we want to find the velocity.
By plugging the known values such as \(a = 5\), \(\omega = 4\), and \(x = 3\) into the formula, the particle's velocity is calculated as 16 units at the given displacement. This illustrates how velocity changes during different stages of oscillation.

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Most popular questions from this chapter

A mass \(m\) is suspended from a string of length \(l\) and force constant \(k\). The frequency of vibration of the mass is \(f_{1}\). The spring is cut in to two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is \(f_{2}\). Which of the following reaction between the frequencies is correct. [NCERT \(198_{3}\); CPMT 1986; MP PMT 1991] (a) \(f_{1}=\sqrt{2} f_{2}\) (b) \(f_{1}=f_{2}\) (c) \(f_{1}=2 f_{2}\) (d) \(f_{2}=\sqrt{2} f_{1}\)

Two equations of two S.H.M. are \(x=a \sin (\omega t-\alpha)\) and \(y=b \cos (\omega t-\alpha)\). The phase difference between the two is (a) \(0^{\circ}\) (b) \(\infty^{\circ}\) (c) \(90^{\circ}\) (d) \(180^{\circ}\)

The period of a simple pendulum is doubled when [CPMT 1974; MNR 1980 ; AFMC 1995\(]\) (a) Its length is doubled (b) The mass of the bob is doubled (c) Its length is made four times (d) The mass of the bob and the length of the pendulum are doubled

Two particles executes S.H.M. of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. The phase difference between them is (a) \(30^{\circ}\) (b) \(60^{\circ}\) (c) \(90^{\circ}\) (d) \(120^{\circ}\)

A particle of mass \(200 \mathrm{gm}\) executes S.H.M. The restoring force is provided by a spring of force constant \(80 \mathrm{~N} / \mathrm{m}\). The time period of oscillations is (a) \(0.31 \mathrm{sec}\) (b) \(0.15 \mathrm{sec}\) (c) \(0.05\) sec (d) \(0.02 \mathrm{sec}\)
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