Velocity of a particle of mass 3 gram performing SHM while crossing its mean position is 6cm/s. If amplitude of SHM is 2cm then the force constant is:
A. 4.5 dyne/cm
B. 9 dyne/cm
C. 27 dyne/cm
D. 1.5 dyne/cm
Answer
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Hint: We are given the mass, amplitude and the mean velocity of a particle in SHM. We have an equation that relates angular frequency and force constant. Using the equation of mean velocity we can find the angular frequency and thus by substituting it in the equation with force constant, we will get the required solution.
Formula used: ${{V}_{mean}}=A\omega $
$\omega =\sqrt{\dfrac{k}{m}}$
Complete step by step answer:
In the question we are given the mass and velocity of a particle performing simple harmonic motion while the particle is crossing the mean position.
Let ‘m’ be the mass of the particle and ‘${{v}_{mean}}$’ be the meanvelocity. Then we are given,
$m=3g$
${{v}_{mean}}=6cm/s$
We are also given the amplitude of the SHM,
$A=2cm$
We know that the velocity of a particle at the mean position in simple harmonic motion is given by the equation,
${{V}_{mean}}=A\omega $, were ‘A’ is the amplitude and ‘$\omega $’ is the angular frequency.
By substituting the values of mean velocity and amplitude in the above equation, we get
$\Rightarrow 6=2\omega $
By solving the above equation, we get the angular frequency as,
$\Rightarrow \omega =\dfrac{6}{2}=3rad/s$
In the question we are asked to find the force constant.
We know that the equation for angular frequency in SHM is given as,
$\omega =\sqrt{\dfrac{k}{m}}$, were ‘k’ is the force constant and ‘m’ is the mass of the particle.
Since we know angular frequency and mass, we can substitute them in the above equation.
$\Rightarrow 3=\sqrt{\dfrac{k}{3}}$
By squaring on both sides of the equation and solving it we will get the force constant as,
$\Rightarrow {{3}^{2}}=\dfrac{k}{3}$
$\Rightarrow k=9\times 3$
$\Rightarrow k=27dyne/cm$
Therefore we get the force constant as 27 dyne/cm.
So, the correct answer is “Option C”.
Note: A simple harmonic motion or SHM is an oscillatory periodic motion, where the restoring force of the moving body is directly proportional to the magnitude of displacement of the body.
Force constant can be defined as the force applied on the body, when the displacement of the body is unity.
Formula used: ${{V}_{mean}}=A\omega $
$\omega =\sqrt{\dfrac{k}{m}}$
Complete step by step answer:
In the question we are given the mass and velocity of a particle performing simple harmonic motion while the particle is crossing the mean position.
Let ‘m’ be the mass of the particle and ‘${{v}_{mean}}$’ be the meanvelocity. Then we are given,
$m=3g$
${{v}_{mean}}=6cm/s$
We are also given the amplitude of the SHM,
$A=2cm$
We know that the velocity of a particle at the mean position in simple harmonic motion is given by the equation,
${{V}_{mean}}=A\omega $, were ‘A’ is the amplitude and ‘$\omega $’ is the angular frequency.
By substituting the values of mean velocity and amplitude in the above equation, we get
$\Rightarrow 6=2\omega $
By solving the above equation, we get the angular frequency as,
$\Rightarrow \omega =\dfrac{6}{2}=3rad/s$
In the question we are asked to find the force constant.
We know that the equation for angular frequency in SHM is given as,
$\omega =\sqrt{\dfrac{k}{m}}$, were ‘k’ is the force constant and ‘m’ is the mass of the particle.
Since we know angular frequency and mass, we can substitute them in the above equation.
$\Rightarrow 3=\sqrt{\dfrac{k}{3}}$
By squaring on both sides of the equation and solving it we will get the force constant as,
$\Rightarrow {{3}^{2}}=\dfrac{k}{3}$
$\Rightarrow k=9\times 3$
$\Rightarrow k=27dyne/cm$
Therefore we get the force constant as 27 dyne/cm.
So, the correct answer is “Option C”.
Note: A simple harmonic motion or SHM is an oscillatory periodic motion, where the restoring force of the moving body is directly proportional to the magnitude of displacement of the body.
Force constant can be defined as the force applied on the body, when the displacement of the body is unity.
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