Physics Oscillations

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New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

Let x = A sin ω t  & v = A ω c o s ω t

v = ω A 2 x 2

v 2 = ω 2 x 2 ω 2 A 2

v 2 ω 2 A 2 + x 2 A 2 = 1

New answer posted

3 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Here, K e q = 2 k

T = 2 π m 2 k

f = 1 2 π 2 k m

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

For equilibrium,

d U d r = 0 1 0 α r 1 1 + 5 β r 6 = 0 r = ( 2 α β ) 1 5

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

T = 2 π l g

2 = 2 π 2 g

g = 2 π 2 m / s 2

New answer posted

3 months ago

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P
Payal Gupta

Contributor-Level 10

Since both spring applied force in same direction.

Equivalent spring constant = 4k

T = 2 π m 4 k

T = π m k

New answer posted

3 months ago

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P
Payal Gupta

Contributor-Level 10

Using general equation of SHM :

y=Asin (ωt+? 0)=A2

ωt+? 0=π6, 5π6

At t = 0

? 0=π6, 5π6

Since it is moving in – x direction

? =5π6

New answer posted

3 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

As we know that maximum acceleration will act on the particle when it is at extreme position .

ω=π60.1=10π6

f=mω2x

fm=ω2A=1000π236*0.36=π2=9.87N

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

T = 2 π ? g where t = l 2

T = 2 π t g

T = x 2 T

2 π t 2 g = x 2 2 π t g

1 2 = x 2 x = 2

New answer posted

3 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

T = m ? ^ ω 2

T = m : ( 2 ω ) 2

T = 4 T

New answer posted

3 months ago

0 Follower 2 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

 For SHM, kinetic and potential energies are equal when displacement x = ± A 2  , where A is the amplitude. At this point, total energy is evenly shared between motion and position-dependent restoring forces.

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