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Question: The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is \(10{s^{ - 1}}\...

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10s110{s^{ - 1}}. At, t=0t = 0 the displacement is 5m5m. What is the maximum acceleration? The initial phase is π4\dfrac{\pi }{4}​.
(a)\left( a \right) 500m/s2500m/{s^2}
(b)\left( b \right) 7502m/s2750\sqrt 2 m/{s^2}
(c)\left( c \right) 750m/s2750m/{s^2}
(d)\left( d \right) 5002m/s2500\sqrt 2 m/{s^2}

Explanation

Solution

Hint So this problem can be solved by using the equation of harmonic which is y=Asin(ωt+π4)y = A\sin \left( {\omega t + \dfrac{\pi }{4}} \right)and also by using the velocity formula which is equal tov=Aωcos(ωt+π4)v = A\omega \cos \left( {\omega t + \dfrac{\pi }{4}} \right). So by using both of these and equating we will get the maximum acceleration.
Formula used
The harmonic equation can be represented by
y=Asin(ωt+π4)y = A\sin \left( {\omega t + \dfrac{\pi }{4}} \right)
Velocity,
v=Aωcos(ωt+π4)v = A\omega \cos \left( {\omega t + \dfrac{\pi }{4}} \right)
And acceleration will be equal to
a=Aω2sin(ωt+π4)a = - A{\omega ^2}\sin \left( {\omega t + \dfrac{\pi }{4}} \right)
Here,
yy, will be the harmonic motion
vv, will be the velocity
aa, will be the acceleration
And ω\omega , will be the angular velocity.

Complete Step By Step Solution We have been given the ratio between the maximum acceleration and the maximum velocity. And there is the displacement also given. So we have to just find the maximum acceleration.
Let the equation for the harmonic motion will be represented as
y=Asin(ωt+π4)y = A\sin \left( {\omega t + \dfrac{\pi }{4}} \right)
And also we know,
Velocity is equal to,
v=Aωcos(ωt+π4)v = A\omega \cos \left( {\omega t + \dfrac{\pi }{4}} \right)
And the acceleration of the body will be equal to
a=Aω2sin(ωt+π4)a = - A{\omega ^2}\sin \left( {\omega t + \dfrac{\pi }{4}} \right)
Since we have been given that the ratio of maximum acceleration to maximum velocity in a simple harmonic motion is10s110{s^{ - 1}} which will be represented by ω\omega
Therefore, substituting the value of time
That is t=0t = 0,
Inharmonic motion equation, we get
5=Asin(π4)5 = A\sin \left( {\dfrac{\pi }{4}} \right)
So from here,
A=52\Rightarrow A = 5\sqrt 2
Now, we will find the maximum acceleration and for this we have
Maximum acceleration =Aw2 = A{w^2}
Therefore, substituting the values, we get
52×10×10\Rightarrow 5\sqrt 2 \times 10 \times 10
On solving the above, the final value will be equal to
5002m/s2\Rightarrow 500\sqrt 2 m/{s^2}

Therefore, 5002500\sqrt 2 is the maximum acceleration, and hence, the option DDis the right choice.

Note When harmonic motion occurs harmonic functions are the waveforms of the motions that result. In other cases, for example, various musical instruments, when the player stimulates the instrument in a certain way, multiple wavelengths of response are being stimulated at the same time. In that case, the waveform produced will be a combination of different frequencies.