
A particle of mass $10gm$ is describing \[S.H.M\] along a straight line with period of $2\sec $ and amplitude of $10cm$ . Its kinetic energy when it is at $5cm$ from its equilibrium position is
A. $37.5{\pi ^2}erg$
B. $3.75{\pi ^2}erg$
C. $375{\pi ^2}erg$
D. $0.375{\pi ^2}erg$
Answer
164.1k+ views
Hint: Simple harmonic motion is a motion in which a particle experiences periodic motion (S.H.M). Not all S.H.M.s are periodic motions, but all periodic motions are S.H.M.s. We must first obtain the expression of velocity by differentiating the displacement expression in order to obtain the relationship between kinetic energy and displacement (by looking at the choices) (i.e. the equation of S.H.M).
Formula Used:
1. The velocity at a displacement y is
$v=\omega {\sqrt {{a^2}-{y^2}}}$
2. The formula of Kinetic Energy as,
$K = \dfrac {1}{2} m {v^2}= \dfrac{1}{2}m{\omega ^2}({a^2} - {y^2})$
where m is mass, omega is angular frequency, a is maximum displacement and y is displacement at given time.
Complete step by step solution:
Before starting the solution let, we write all the given things which is needed to solve above equation are,
Mass of the particle(m) = $10gm$
The value of omega, $\omega = \left( {\dfrac{{2\pi }}{2}} \right)$
Amplitude, $a = 10cm$
Distance from the equilibrium, $y = 5cm$
By putting all the values in the formula of kinetic energy we get,
$K = \dfrac{1}{2} \times 10 \times {\left( {\dfrac{{2\pi }}{2}} \right)^2}\left[ {{{\left( {10} \right)}^2} - {{\left( 5 \right)}^2}} \right]$
By solving the above equation, we get the solution of the equation as,
$K = 375{\pi ^2}erg$
As a result, we get the kinetic energy when it is at $5cm$ from its equilibrium position is $K = 375{\pi ^2}erg$ . Therefore, the correct answer is $375{\pi ^2}erg$ .
Hence, the correct option is (C).
Note: Simple harmonic motion is a periodic motion in which the restoring force on the moving item works in the direction of the object's equilibrium position and is directly proportionate to the magnitude of the displacement. The amplitude, mass, and displacement of the item, as well as its angular frequency, all affect the S.H.M.'s kinetic energy.
Formula Used:
1. The velocity at a displacement y is
$v=\omega {\sqrt {{a^2}-{y^2}}}$
2. The formula of Kinetic Energy as,
$K = \dfrac {1}{2} m {v^2}= \dfrac{1}{2}m{\omega ^2}({a^2} - {y^2})$
where m is mass, omega is angular frequency, a is maximum displacement and y is displacement at given time.
Complete step by step solution:
Before starting the solution let, we write all the given things which is needed to solve above equation are,
Mass of the particle(m) = $10gm$
The value of omega, $\omega = \left( {\dfrac{{2\pi }}{2}} \right)$
Amplitude, $a = 10cm$
Distance from the equilibrium, $y = 5cm$
By putting all the values in the formula of kinetic energy we get,
$K = \dfrac{1}{2} \times 10 \times {\left( {\dfrac{{2\pi }}{2}} \right)^2}\left[ {{{\left( {10} \right)}^2} - {{\left( 5 \right)}^2}} \right]$
By solving the above equation, we get the solution of the equation as,
$K = 375{\pi ^2}erg$
As a result, we get the kinetic energy when it is at $5cm$ from its equilibrium position is $K = 375{\pi ^2}erg$ . Therefore, the correct answer is $375{\pi ^2}erg$ .
Hence, the correct option is (C).
Note: Simple harmonic motion is a periodic motion in which the restoring force on the moving item works in the direction of the object's equilibrium position and is directly proportionate to the magnitude of the displacement. The amplitude, mass, and displacement of the item, as well as its angular frequency, all affect the S.H.M.'s kinetic energy.
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