The maximum velocity and the maximum acceleration of a body moving in a simple harmonic oscillator are $2m/s$ and $4m/{{s}^{2}}$. The angular velocity will be:
$A)\text{ }3rad/s$
$B)\text{ 0}\text{.5}rad/s$
$C)\text{ 1}rad/s$
$D)\text{ 2}rad/s$
Answer
601.2k+ views
Hint: This problem can be solved by using the direct formulae for the maximum velocity and maximum acceleration in a simple harmonic oscillator in terms of its amplitude and angular velocity.
Formula used:
$v=A\omega $
$a=A{{\omega }^{2}}$
Complete step-by-step answer:
We will use the formula for the maximum velocity and acceleration of a body in simple harmonic oscillation.
The maximum acceleration $a$ of a body in simple harmonic oscillation of amplitude $A$ and angular frequency $\omega $ is given by
$a=A{{\omega }^{2}}$ --(1)
The maximum velocity $v$ of a body in simple harmonic oscillation of amplitude $A$ and angular frequency $\omega $ is given by
$v=A\omega $ --(2)
Now, let us analyze the question.
Let the amplitude of the body in simple harmonic motion be $A$.
Let the angular frequency of the body in simple harmonic motion be $\omega $.
The maximum acceleration of the body is $a=4m/{{s}^{2}}$.
The maximum velocity of the body is $v=2m/s$.
Therefore, using (1), we get
$a=A{{\omega }^{2}}$ --(3)
Also, using (2), we get
$v=A\omega $ --(4)
Dividing (3) by (4), we get
$\dfrac{a}{v}=\dfrac{A{{\omega }^{2}}}{A\omega }=\omega $ --(5)
Putting the values of the variables in (5), we get
$\omega =\dfrac{a}{v}=\dfrac{4}{2}=2rad/s$
Hence, the required angular velocity of the body is $2rad/s$.
Therefore, the correct answer is $D)\text{ 2}rad/s$.
Note: We could have also found out a relation between the amplitude and the angular frequency from the maximum velocity and then use this relation in the equation for the maximum acceleration to get the angular velocity. However, we eliminated the amplitude in one step by dividing the equation for the angular acceleration with the angular velocity. We did this so that we can make the calculation shorter and eliminate the usage of the unnecessary variable, that is, the amplitude.
Formula used:
$v=A\omega $
$a=A{{\omega }^{2}}$
Complete step-by-step answer:
We will use the formula for the maximum velocity and acceleration of a body in simple harmonic oscillation.
The maximum acceleration $a$ of a body in simple harmonic oscillation of amplitude $A$ and angular frequency $\omega $ is given by
$a=A{{\omega }^{2}}$ --(1)
The maximum velocity $v$ of a body in simple harmonic oscillation of amplitude $A$ and angular frequency $\omega $ is given by
$v=A\omega $ --(2)
Now, let us analyze the question.
Let the amplitude of the body in simple harmonic motion be $A$.
Let the angular frequency of the body in simple harmonic motion be $\omega $.
The maximum acceleration of the body is $a=4m/{{s}^{2}}$.
The maximum velocity of the body is $v=2m/s$.
Therefore, using (1), we get
$a=A{{\omega }^{2}}$ --(3)
Also, using (2), we get
$v=A\omega $ --(4)
Dividing (3) by (4), we get
$\dfrac{a}{v}=\dfrac{A{{\omega }^{2}}}{A\omega }=\omega $ --(5)
Putting the values of the variables in (5), we get
$\omega =\dfrac{a}{v}=\dfrac{4}{2}=2rad/s$
Hence, the required angular velocity of the body is $2rad/s$.
Therefore, the correct answer is $D)\text{ 2}rad/s$.
Note: We could have also found out a relation between the amplitude and the angular frequency from the maximum velocity and then use this relation in the equation for the maximum acceleration to get the angular velocity. However, we eliminated the amplitude in one step by dividing the equation for the angular acceleration with the angular velocity. We did this so that we can make the calculation shorter and eliminate the usage of the unnecessary variable, that is, the amplitude.
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