a particle is executing SHM with time period of 4 second if its amplitude is 5 cm calculate displacement velocity and acceleration of the particle after 1 second it starts oscillating
Answer (1)
9th Nov, 2020
Hello!
I will help you with the solution to this question.
We have, t= 4s and a=5 cm
Displacement(y)= a sinwt
=>y = 5sin(2π/4)t = 5sin(πt/2)
Also, v= dy/dt
=> v= (d/dt)[5sin(πt/2)] = (5/2)[πcos(πt/2)]
Again, a= dv/dt
=> a=( -5π^2/2)[sin(πt/2)]
After 1s,
y= 5sin(π/2×1) = 5sin(π/2) = 5cm
v= 5πcos(πt/2) = 5×(22/7)× cos(π/2×1)= 0 cm/s
a=( -5π^2/4)sin(π/2×1) = - 5π^2/4 cm/s^2 .
If you find the above solution a little confusing due to the text, I am giving the google drive link of my handwritten answer(copy and paste the link in your browser to view it):-
https://drive.google.com/file/d/19b3hb9YR8OVzdIzI9eKV2PnU6cYA5nSl/view?usp=drivesdk
Hope I was able to help you!
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Comments (9)
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please explain the acceleration part again
9th Nov, 2020
Svvvvvv
Do not directly click on 'web search' option....copy the link and then go and paste in google chrome browser...the link will open fine.
9th Nov, 2020
Svvvvvv
The acceleration is derivative of velocity, so I found out derivate of the velocity to find it out and used the given condition after 1 s.
I am unable to paste it even
please can you again solve the acceleration part
can you solve 5π^2/4
9th Nov, 2020
Svvvvvv
That is the final answer...if you want a value put π=22/7 or π=3.14 simple. You will get your answer.
Well after calculating the acceleration is coming -12.32cm/s^2
Well after calculating the acceleration is coming -12.32cm/s^2
9th Nov, 2020
Svvvvvv
I don not know why are you not able to paste it in your browser..it is so simple. If you cannot do this, then I cannot help you more.
