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# if the mass of the pulleys shown in figure are small and the cord is inextensible, the angular frequency of oscillation of the system is

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Question

If the mass of the pulleys shown in figure is very very small and the cord is inextensible, the angular frequency of oscillation of the system is

Video Solution Open in App Solution

The correct option is D

√ k a k b 4 m ( k a + k b ) Let T

be the tension in the cord and

x a and x b

the displacement of pulleys

A and B respectively.

Now assume that pulley

A

is fixed; the extension in spring will be

x b = x 2

, as the block gets lowered by

x ⇒ x = 2 x b

Similarly if we imagine that pulley

B is fixed, x = 2 x a .

However neither pulley

B nor pulley A is fixed. ⇒ x = 2 x a + 2 x b . . . ( 1 )

From balancing force on pulleys( pulley is massless)

2 T = k b x b . . . ( 2 ) and 2 T = k a x a . . . ( 3 ) If k e q

denotes equivalent spring constant, for the spring block system.

⇒ T = k e q x ⇒ T k e q = x ⇒ T k e q = 2 x a + 2 x b . . . . ( 4 ) From equations ( 2 ) and ( 3 ) , ( 4 ) x a = 2 T k a and x b = 2 T k b and ⇒ k e q = 1 4 ( 1 k a + 1 k b )

Hence angular frequency of oscillation is,

ω = √ k e q m = √ k a k b 4 m ( k a + k b )

SHM in Vertical Spring Block

Standard XII Physics

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## If the mass of the pulley shown in figure is small and the cord is inextensible, the angular 1of oscillation of the system is :

Click here👆to get an answer to your question ✍️ If the mass of the pulley shown in figure is small and the cord is inextensible, the angular 1of oscillation of the system is :

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>>Oscillations Due to Spring

>>If the mass of the pulley shown in figur

If the mass of the pulley shown in figure is small and the cord is inextensible, the angular 1

Question

of oscillation of the system is :

A

m K a ​ +K b ​ ​ ​

B

(K a ​ +K b ​ )m K a ​ K b ​ ​ ​

C

4m(K a ​ +K b ​ ) K a ​ K b ​ ​ ​

D

(K a ​ +K b ​ ).m 4K a ​ K b ​ ​ ​ Hard Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is A)

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## A block of mass m is attached to one end of a light inextensible string passing over a smooth light pulley B and under another smooth light pulley A as shown in the figure. The other end of the string is fixed to a ceiling. A and B are held by spring of spring constants k(1) and k(2). Find angular frequency of small oscillations of small oscillations of the system.

Let T be the tension in the cord and x(a) and x(b) the displacements of pulleys A and b respectively. Now assume that pulley B is fixed, then extension of spring x(b) = (x)/(2) or x = 2x(b). Similarly if we imagine that pulley A is fixed, x = 2x(a).But neither pulley B nor pulley A is fixed. :. x = 2x(a) +2x(b) ....(1) From free body diagram of pulleys, 2T = k(b)x(b) ...(2) or 2T = k(a)x(a) ...(3) If k(eq) denotes equivalent spring constant, (T)/(k(eq)) =x =2x(a) +2x(b) From equations (2) and (3) x(a) = (2T)/(k(a)) and x(b) = (2T)/(k(b)) i.e., k(eq) = (1)/(4((1)/(k(a))+(1)/(k(b)))) Hence omega = sqrt((k(eq))/(m)) = sqrt((k(a)k(b))/(4m(k(a)+k(b)))) 2nd Method

Home > English > Class 11 > Physics > Chapter > Oscillations >

A block of mass m is attached ...

A block of mass m m

is attached to one end of a light inextensible string passing over a smooth light pulley

B B

and under another smooth light pulley

A A

as shown in the figure. The other end of the string is fixed to a ceiling.

A A and B B

are held by spring of spring constants

k 1 k1 and k 2 k2

. Find angular frequency of small oscillations of small oscillations of the system.

Updated On: 27-06-2022

00 : 30

Text Solution Open Answer in App A K a + K b m − − − − − − − − √ Ka+Kbm B K a K b ( K a + K b )m − − − − − − − − − − − √ KaKb(Ka+Kb)m C K a K b 4m( K a + K b ) − − − − − − − − − − − − √ KaKb4m(Ka+Kb) D 4 K a K b ( K a + K b )m − − − − − − − − − − − √ 4KaKb(Ka+Kb)m Answer

Solution Let T T

be the tension in the cord and

x a xa and x b xb

the displacements of pulleys

A A and b b

respectively. Now assume that pulley

B B

is fixed, then extension of spring

x b = x 2 xb=x2 or x=2 x b x=2xb

. Similarly if we imagine that pulley

A A is fixed, x=2 x a x=2xa .But neither pulley B B nor pulley A A is fixed. ∴x=2 x a +2 x b ....(1) ∴x=2xa+2xb....(1)

From free body diagram of pulleys,

2T= k b x b ...(2) 2T=kbxb...(2) or 2T= k a x a ...(3) 2T=kaxa...(3) If k eq keq

denotes equivalent spring constant,

T k eq =x=2 x a +2 x b Tkeq=x=2xa+2xb

From equations (2) and (3)

x a = 2T k a xa=2Tka and x b = 2T k b i.e., k eq = 1 4( 1 k a + 1 k b )

xb=2Tkbi.e.,keq=14(1ka+1kb)

Hence ω= k eq m − − − √ = k a k b 4m( k a + k b ) − − − − − − − − − − − √

ω=keqm=kakb4m(ka+kb)

2nd Method

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Mohammed 6 day ago

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