shm


Options are:
(a) √9k/m
(b) √8k/m
(c) √k/m
(d) √k/2m

33 Answers

33
Abhishek Priyam ·

yup..

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

33
Abhishek Priyam ·

dont worry about pulley's friction.. :p

33
Abhishek Priyam ·

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

106
Asish Mahapatra ·

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......

33
Abhishek Priyam ·

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

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

33
Abhishek Priyam ·

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

106
Asish Mahapatra ·

lekin pulley to frictionless hai naa, to phir ghuumega kaise?

33
Abhishek Priyam ·

are yaar...

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

106
Asish Mahapatra ·

thread can slip but tension has to remain the same.... is that right?

106
Asish Mahapatra ·

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

106
Asish Mahapatra ·

thx abhi

33
Abhishek Priyam ·

:)

33
Abhishek Priyam ·

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

62
Lokesh Verma ·

awesome priyam :)

33
Abhishek Priyam ·

Sky's solution kyun nahi pink kiye...

62
Lokesh Verma ·

arrey mujhe laga tha that was ur post... (Din see it was sky's :)

3
iitimcomin ·

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..........

39
Dr.House ·

the two springs are in series. so effective spring constant is 9k. therefore answer is a ...............

106
Asish Mahapatra ·

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

1
skygirl ·

Fr = -kx !

how meter2 ??

33
Abhishek Priyam ·

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]

1
skygirl ·

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

106
Asish Mahapatra ·

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

1
skygirl ·

ps: dun think abt series n parallel .. solve it simply from basics..

1
skygirl ·

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]

33
Abhishek Priyam ·

kk... just a sec

106
Asish Mahapatra ·

abhi pls explain

33
Abhishek Priyam ·

never be dependent on series parallel in springs...

33
Abhishek Priyam ·

(b)

correct..

33
Abhishek Priyam ·

interesting...

both type of combi in above posts..

seires by mathe.. parallel by rkrish..

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