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

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....

thx abhi

:)

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

Sky's solution kyun nahi pink kiye...

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 ]

2x = Fr/K1 + Fr/K2

=> x = Fr [1/k1 + 1/k2 ] /2

but 2Fr =mw²r [y?[1]]

so, x = 2Fr [1/k1 + 1/k2 ]/4

=> x = mw²x [1/k1 + 1/k2 ] /4

=> w² = 4k1k2/(k1+k2)/m

here, k1=6k, k2=3k , m=m :P ,

so, w² = 8k/m [1]

never be dependent on series parallel in springs...

interesting...

both type of combi in above posts..

seires by mathe.. parallel by rkrish..