
Mark the wrong statement about Simple Harmonic motion (S.H.M)
(A) All S.H.M.s have fixed time period
(B) All motion having same time period are S.H.M.
(C) In S.H.M. total energy is proportional to square of amplitude
(D) Phase constant of S.H.M. depends upon initial conditions
Answer
232.8k+ views
Hint: First start with the basic knowledge about simple harmonic motion then find the formula for the force, time period, total energy and phase constant in the case of particles moving in the simple harmonic motion. Use the elimination method for all the options mentioned above in the question and you will get the required answer.
Complete answer:
First start with the Option(A):
In S.H.M, we know the force on the particle is as follows;
$F = - kx$
Also we know, $F = ma$
From above two equations, we get;
$a = - \dfrac{k}{m}x$
Another form of acceleration, $a = - {\omega ^2}x$
Equating both the equation of acceleration, we get;
${\omega ^2} = \dfrac{k}{m}$
$\omega = \sqrt {\dfrac{k}{m}} $ where both k and m are constant hence w is also constant.
We know, time period is given by;
$T = \dfrac{{2\pi }}{\omega }$ , as $\omega $is constant here hence the time period will also be constant.
Therefore, option(A) is correct.
Now, for option(B):
When a particle is in a circular motion its time period is also constant but it is not S.H.M.
Hence all motion having same time period are not S.H.M.
Therefore, option(B) is a wrong statement.
Now, for option(C):
In S.H.M, total energy $T.E = \dfrac{1}{2}K{A^2}$ , where A is amplitude.
Hence option(C) is the correct statement.
Now, for option(D):
S.H.M is represented as;
$x = A\sin \left( {\omega t + \Phi } \right)$ , where $\Phi ,is\,phase\,constant$
Here phase constant depends on the initial position, therefore option(D) is correct statement.
Hence the correct answer is Option(B).
Note: In this type of question you need to use the method of elimination in order to get the right answer. Start with option 1 and move to all option. Know the basic formulas of S.H.M and use all the formula accurately and use all the information given in the carefully and finally you will get the required answer.
Complete answer:
First start with the Option(A):
In S.H.M, we know the force on the particle is as follows;
$F = - kx$
Also we know, $F = ma$
From above two equations, we get;
$a = - \dfrac{k}{m}x$
Another form of acceleration, $a = - {\omega ^2}x$
Equating both the equation of acceleration, we get;
${\omega ^2} = \dfrac{k}{m}$
$\omega = \sqrt {\dfrac{k}{m}} $ where both k and m are constant hence w is also constant.
We know, time period is given by;
$T = \dfrac{{2\pi }}{\omega }$ , as $\omega $is constant here hence the time period will also be constant.
Therefore, option(A) is correct.
Now, for option(B):
When a particle is in a circular motion its time period is also constant but it is not S.H.M.
Hence all motion having same time period are not S.H.M.
Therefore, option(B) is a wrong statement.
Now, for option(C):
In S.H.M, total energy $T.E = \dfrac{1}{2}K{A^2}$ , where A is amplitude.
Hence option(C) is the correct statement.
Now, for option(D):
S.H.M is represented as;
$x = A\sin \left( {\omega t + \Phi } \right)$ , where $\Phi ,is\,phase\,constant$
Here phase constant depends on the initial position, therefore option(D) is correct statement.
Hence the correct answer is Option(B).
Note: In this type of question you need to use the method of elimination in order to get the right answer. Start with option 1 and move to all option. Know the basic formulas of S.H.M and use all the formula accurately and use all the information given in the carefully and finally you will get the required answer.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

