
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
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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.
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