
A body oscillates with SHM according to the equation (in S.I. unit) $x = \cos [2\pi t + \dfrac{\pi }{4}]$. At $t = 1.5\sec $, calculate the
(1) Displacement
(2) Speed and
(3) Acceleration of the body
Answer
124.8k+ views
Hint: Simple harmonic motion is a special type of periodic motion. In this type of motion the acceleration of the particle is directly proportional to the displacement of the body from its mean position. The force is always directed towards the mean position. It is a special case of oscillatory motion.
Complete step by step solution:
1. Displacement is defined as the change in the position of the object.
Given $x = \cos [2\pi t + \dfrac{\pi }{4}]$---(i)
At $t = 1.5\sec $
Substituting the value of time in the equation (i),
$\Rightarrow {x_{t = 1.5}} = \cos [2\pi \times 1.5 + \dfrac{\pi }{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos [3\pi + \dfrac{\pi }{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos [\dfrac{{7\pi + \pi }}{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos \dfrac{{8\pi }}{4}$
$\Rightarrow {x_{t = 1.5}} = \cos [2\pi ]$
$\cos n\pi = 1$ if n is an even number.
$\therefore $ ${x_{t = 1.5}} = 1m$
Displacement of the particle is $x = 1m$
2. Speed is the rate of change of displacement with time. It can be writ
$ \Rightarrow v = \dfrac{{dx}}{{dt}}$
$\Rightarrow v = \dfrac{d}{{dt}}[\cos (2\pi t + \dfrac{\pi }{4})]$
$\Rightarrow v = - \sin (2\pi t + \dfrac{\pi }{4}).2\pi $---(ii)
At $t = 1.5\sec $
Substituting the value of time in equation (ii)
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (2\pi \times 1.5 + \dfrac{\pi }{4})$
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (3\pi + \dfrac{\pi }{4})$
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (2\pi )$
$\sin n\pi = 0$ if n is an even number
$\therefore {v_{t = 1.5}} = 0$
The speed of the particle is $v = 0$.
3. Acceleration is the rate of change of velocity with respect to time. It can be written by using the formula
$a = \dfrac{{dv}}{{dt}}$
$\Rightarrow a = \dfrac{d}{{dt}}[ - 2\pi \sin (2\pi t + \dfrac{\pi }{4})]$
$\Rightarrow a = - 2\pi \cos (2\pi t + \dfrac{\pi }{4}) \times 2\pi $---(iii)
At $t = 1.5\sec $
Substituting the value of time in equation (iii)
$\Rightarrow a = - 2\pi \cos (2\pi \times 1.5 + \dfrac{\pi }{4}) \times 2\pi $
$\Rightarrow a = - 4{\pi ^2}\cos (3\pi + \dfrac{\pi }{4})$
$\Rightarrow a = - 4{\pi ^2}\cos (2\pi )$
$\cos n\pi = 1$if n is an even integer
$\therefore a = - 4{\pi ^2}$
The acceleration of the particle is $a = - 4{\pi ^2}$.
Note: It is to be noted that though simple harmonic motion is a special case of oscillatory motion, they are different. Oscillatory motion is the to and fro motion of the particle from its mean position. An object can be in oscillatory motion forever in the absence of friction. But simple harmonic motion is a type of oscillatory motion in which the particle moves along a straight line such that magnitude is proportional to the distance.
Complete step by step solution:
1. Displacement is defined as the change in the position of the object.
Given $x = \cos [2\pi t + \dfrac{\pi }{4}]$---(i)
At $t = 1.5\sec $
Substituting the value of time in the equation (i),
$\Rightarrow {x_{t = 1.5}} = \cos [2\pi \times 1.5 + \dfrac{\pi }{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos [3\pi + \dfrac{\pi }{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos [\dfrac{{7\pi + \pi }}{4}]$
$\Rightarrow {x_{t = 1.5}} = \cos \dfrac{{8\pi }}{4}$
$\Rightarrow {x_{t = 1.5}} = \cos [2\pi ]$
$\cos n\pi = 1$ if n is an even number.
$\therefore $ ${x_{t = 1.5}} = 1m$
Displacement of the particle is $x = 1m$
2. Speed is the rate of change of displacement with time. It can be writ
$ \Rightarrow v = \dfrac{{dx}}{{dt}}$
$\Rightarrow v = \dfrac{d}{{dt}}[\cos (2\pi t + \dfrac{\pi }{4})]$
$\Rightarrow v = - \sin (2\pi t + \dfrac{\pi }{4}).2\pi $---(ii)
At $t = 1.5\sec $
Substituting the value of time in equation (ii)
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (2\pi \times 1.5 + \dfrac{\pi }{4})$
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (3\pi + \dfrac{\pi }{4})$
$\Rightarrow {v_{t = 1.5}} = - 2\pi \sin (2\pi )$
$\sin n\pi = 0$ if n is an even number
$\therefore {v_{t = 1.5}} = 0$
The speed of the particle is $v = 0$.
3. Acceleration is the rate of change of velocity with respect to time. It can be written by using the formula
$a = \dfrac{{dv}}{{dt}}$
$\Rightarrow a = \dfrac{d}{{dt}}[ - 2\pi \sin (2\pi t + \dfrac{\pi }{4})]$
$\Rightarrow a = - 2\pi \cos (2\pi t + \dfrac{\pi }{4}) \times 2\pi $---(iii)
At $t = 1.5\sec $
Substituting the value of time in equation (iii)
$\Rightarrow a = - 2\pi \cos (2\pi \times 1.5 + \dfrac{\pi }{4}) \times 2\pi $
$\Rightarrow a = - 4{\pi ^2}\cos (3\pi + \dfrac{\pi }{4})$
$\Rightarrow a = - 4{\pi ^2}\cos (2\pi )$
$\cos n\pi = 1$if n is an even integer
$\therefore a = - 4{\pi ^2}$
The acceleration of the particle is $a = - 4{\pi ^2}$.
Note: It is to be noted that though simple harmonic motion is a special case of oscillatory motion, they are different. Oscillatory motion is the to and fro motion of the particle from its mean position. An object can be in oscillatory motion forever in the absence of friction. But simple harmonic motion is a type of oscillatory motion in which the particle moves along a straight line such that magnitude is proportional to the distance.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Mark the correct statements about the friction between class 11 physics JEE_Main

Find the acceleration of the wedge towards the right class 11 physics JEE_Main

A standing wave is formed by the superposition of two class 11 physics JEE_Main

Derive an expression for work done by the gas in an class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
