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01:00 - 14:0014:00 - 18:00

Hema
Hema
14:00
user image
It is similar to this image but there is a vertical wall to the right as well a distance A from the mass
sammy gerbil
sammy gerbil
Ah yes the mass makes contact with a wall on the right and loses KE.
Hema
Hema
The spring is first unextended but then is compressed by 2A and released
@sammygerbil yes, actually I'm having trouble uploading images so I took one from google images
sammy gerbil
sammy gerbil
The KE just before the block makes contact with the wall = initial elastic energy - final elastic energy = $\frac12 k[(2A)^2-A^2]$.
Hema
Hema
@sammygerbil ohhh ok
sammy gerbil
sammy gerbil
So next you can work out the KE just after the collision, and that equals the elastic energy stored when the block next comes to rest after the collision.
Hema
Hema
14:08
@sammygerbil by that we can find the new amplitude?
sammy gerbil
sammy gerbil
Yes, because the equilibrium position is always where the extension of the spring is zero (ie its natural length).
Hema
Hema
@sammygerbil how do I get the time taken?
sammy gerbil
sammy gerbil
So the amplitude of oscillation before collision is $2A$.
Hema
Hema
@sammygerbil ohhh ok
sammy gerbil
sammy gerbil
You get time from the equation of motion $x=A\cos(\omega t)$.
Hema
Hema
14:11
@sammygerbil ohhh ok ok
Thank you!
sammy gerbil
sammy gerbil
14:22
@Abcd chat.stackexchange.com/transcript/message/46067902#46067902 This can be solved using the conservation of energy equation. See Finding the period of an anharmonic oscillation by substituting the solution for SHM
@Abcd Alternatively Period $T$ of oscillation with cubic force function
@Hema chat.stackexchange.com/transcript/message/46077259#46077259 Yes this is correct.
Abcd
Abcd
15:03
@sammygerbil Wow! Amazing method. Thanks
@sammygerbil The message doesn't get displayed if along with the link some other words are typed too
 
2 hours later…
sammy gerbil
sammy gerbil
17:16
@Abcd Thanks I wondered what I was doing wrong.
01:00 - 14:0014:00 - 18:00

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