Doubt from shm

2 Likes

I will say the concept I thought . When pan is pulled a little the water loses contact at one side ( so no normal reaction in one side ) and the water suddenly occupies less volume now (2l-x)*h .
The normal reaction at another side now acts as restoring force .

It’s like water is suddenly compressed .Now this restoring force is proportional to the change in volume ( so SHM )

Have they given base length ? If u find anything wrong say me


I try more .. if I get I send...source of question?

Ans is 3 this question is asked in previous year AITS

1 Like

S aits 2018 ... can u send ur solution

Ignore the above trial of mine ...I misunderstood flat with horizontal

1 Like

I don't know how they find (y_2 -y_1)=\frac{x_0^2}{6h} @Bhuvanitha_2020

Got now

Can you explain the restoring part (How they calculated F_R )@Bhuvanitha_2020

I couldn't got what they had done so I posted it

Can you explain what they had done?
@Satyendra_2019 @Bhuvanitha_2020

Don't know about restoring force
But in first two steps they try to find the equation COM
That is if you displace the pan by x distance by how much height the COM goes up during sloshing @akash_2020

2 Likes

Even it didn’t make sense to me completely . The com experience only gravity so parabola trajectory it will follow ..I will try..

2 Likes

I got it posting in a minute

\displaystyle \frac{dU}{dx}=-\frac{3h}{L^2}mg(x-x_1)
As at ∆x=0 ,F_R=0 so we get
\displaystyle ma=-\frac{3h}{L^2}mg(∆x)

2 Likes

∆x is displacement of COM along x axis from equilibrium position(x_1)
Note that restoring force F_R(x_1)=0

2 Likes

Thanks @Satyendra_2019 and @Bhuvanitha_2020

2 Likes