
A particle performing SHM has a maximum velocity $20cm/s$ and maximum acceleration $80cm/{s^2}$ . Its amplitude will be:
(A) $10cm$
(B) $2cm$
(C) $5cm$
(D) $8cm$
Answer
139.5k+ views
Hint The maximum velocity of a particle in simple harmonic motion is given by $v = r\omega $ and the maximum acceleration of a particle in simple harmonic motion is given by $a = r{\omega ^2}$ . Divide one by the other and find the value $\omega $ . Use this value in either of the equations to find the value of the amplitude of the particle in simple harmonic motion.
Complete Step by step answer
Let the maximum velocity of the particle in simple harmonic motion be $v$ . Let the maximum acceleration of the particle be $a$ . Let the amplitude of the particle be $r$ . Let $\omega $ be the frequency of the particle executing simple harmonic motion. It is given in the question that the maximum velocity of the particle is $20cm/s$ and the maximum acceleration of the particle is $80cm/{s^2}$ .
The maximum velocity of a particle executing simple harmonic motion is given by
$v = r\omega $
Substituting the value of maximum velocity, we get
$20 = r\omega $
The maximum acceleration of a particle executing simple harmonic motion is given by
$a = r{\omega ^2}$
Substituting the maximum value of acceleration, we get
$80 = r{\omega ^2}$
Dividing the above equation for maximum acceleration by the already written equation for maximum velocity, we get
$\dfrac{{80}}{{20}} = \omega $
$ \Rightarrow \omega = 4Hz$
Substituting this value of frequency into the equation for maximum velocity, we get
$20 = r \times 4$
By cross multiplication, we get
$r = \dfrac{{20}}{4}$
$ \Rightarrow r = 5cm$
That is, the maximum amplitude calculated is $5cm$ .
Hence, option (C) is the correct option.
Note
Simple harmonic motion is a type of periodic motion. In simple harmonic motion, the amplitude of a particle will also be its maximum displacement from the mean position of the particle. The particle will have maximum velocity when it is at the mean position and the particle will have maximum acceleration when it is at its extreme positions.
Complete Step by step answer
Let the maximum velocity of the particle in simple harmonic motion be $v$ . Let the maximum acceleration of the particle be $a$ . Let the amplitude of the particle be $r$ . Let $\omega $ be the frequency of the particle executing simple harmonic motion. It is given in the question that the maximum velocity of the particle is $20cm/s$ and the maximum acceleration of the particle is $80cm/{s^2}$ .
The maximum velocity of a particle executing simple harmonic motion is given by
$v = r\omega $
Substituting the value of maximum velocity, we get
$20 = r\omega $
The maximum acceleration of a particle executing simple harmonic motion is given by
$a = r{\omega ^2}$
Substituting the maximum value of acceleration, we get
$80 = r{\omega ^2}$
Dividing the above equation for maximum acceleration by the already written equation for maximum velocity, we get
$\dfrac{{80}}{{20}} = \omega $
$ \Rightarrow \omega = 4Hz$
Substituting this value of frequency into the equation for maximum velocity, we get
$20 = r \times 4$
By cross multiplication, we get
$r = \dfrac{{20}}{4}$
$ \Rightarrow r = 5cm$
That is, the maximum amplitude calculated is $5cm$ .
Hence, option (C) is the correct option.
Note
Simple harmonic motion is a type of periodic motion. In simple harmonic motion, the amplitude of a particle will also be its maximum displacement from the mean position of the particle. The particle will have maximum velocity when it is at the mean position and the particle will have maximum acceleration when it is at its extreme positions.
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