Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
can you tell me the problem number 30 in shm chapter in hc verma part 1 ....even i see the solution i am not able tp understand please explain......
SOLUTION
Let the amplitude of oscillation of m and M be x1 and x2 respectievely
a) From the law of conservation of momentum :-
mx1 = Mx2 ...........(1) [bcoz internal forces are present]
Again , 1/2kx02 = 1/2k (x1 + x2)2
Therefore x0 = x1 + x2
[Block and mass oscillates in opposite direction. But x----> stretched part]
From equations (1) and (2) :-
x0 =x1 +(m/M)x1 = [(m+M)/M]x1
Therefore x1(Mx0/m+M)
So x2 = x0 -x1
= mx0/m+M
b) At any positions , let v1 and v2 be the velocities
where v1 = velocity of m w.r.t M
By energy method :-
Total Energy = Constant
(1/2)Mv2 + 1/2m(v1 - v2)2 +1/2k(x1 - x2)2 = constant ........(1)
By using the law of conservation of momentum we get :-
x1 = (M/m)x2 .............(2)
v1 - v2 = (M/m)v2...................(3)
Putting the values in (1) we get:-
mv22 + k(1 + M/m)x22 = constant
Taking derivative from both sides we get :-
w=√[k(m+M)/Mm]
Therefore Time period will be 2∏w
On putting the value of w you will get the answer.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !