shm

yup..

T has to remain same... and thers no problem in slipping.. :)

dont worry about pulley's friction.. :p

that segment m=0 is any segment of rope...

if rope is same tension will be same throughout the rope for massless and friction condition

but priyam in the question, if the extensions in the springs will not be equal, then the system will become tilted. thats y im thinking that extensions in both springs will be same......

tilt kaise hoga... pulley ghum jaega... i was fool.. :P

thread will slip over pulley.. ...edited..

here pulley is not rotating just thread slipping over it...as pointed by Ashish in below post.. :)

are yaar...

pulley nahi...
thread uspar slip kar jaega....

@sky: pls ignore it.... i was out of my mind then....

[3] But is thread ka painting kyun nahi hua... [3]

simple 1........
if u displace block bt x then springs will get displaced by 2x......................
so net tension = T...................
T = 6k(w)....T/2=3kw.......
T=3Kj..............
3/2T=3K(w+j)....w+j=2x...........
T=4kx......
but 2T ACTS ON BODY!!!!!!!!!
2T=8kx.......................
mw^2x = 8kx ........
w=√8k/m..........

priyam : why tension is same in both threads? i might be asking silly questions but still please answer it..

Fr = -kx !

how meter² ??

Let x₁ and x₂ be elongation in 3K and 6K springs

now since tension in thread is same above pulley so 6Kx₂=3Kx₁
x₁=2x₂=T

Now work done by tension is zero.. so
Tx₁+Tx₂=2Tx.... where x is displacement of block...

solving these eqns we get..

x₁=4x/3

so T=3Kx₁=4Kx

but force on block is... 2T or 8kx=ma so ω=√8k/m

Why i have neglected gravity... [4]

well forgot to tell.. Fr = restoring force... [hope this is understood .. but still ]

sky y have u done Fr/k1 btw Fr/k1 has unit meter^2 where as LHS has unit meter.

abhi pls explain

never be dependent on series parallel in springs...

(b)

correct..

interesting...

both type of combi in above posts..

seires by mathe.. parallel by rkrish..