Physics Oscillations

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Range R = u 2 s i n 2 θ g

For 42° and 48° Range will be same

H m a x α  maximum q

So maximum height will be for 48°

 

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Maximum energy = 10 J

1 2 K x 2 = 1 0              

K = 5

Given Tpendulum = Tspring

2 π l g = 2 π m K             

4 g = 5 5           

g = 4m/s2

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Time period T = 2π l g e f f

T i = T = 2 π l g    

T f = T ' = 2 π 4 l / 3 g ( 1 ρ σ ) = 2 π 1 6 l 9 g

T ' = 4 3 T

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

In SHM sum of kinetic and potential energy will be constant and average kinetic energy & average potential energy in one time will be remains same.

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

x = A s i n ω t   (Eq. of a particle executing SHM)

When KE = PE

1 2 m v 2 = 1 2 k x 2

1 2 m ω 2 ( A 2 x 2 ) = 1 2 k x 2 [ ? k = m ω 2 ]

A2 – x2 = x2

  S o , A 2 = A s i n ω t

1 2 = s i n ω t

t = T 8

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Velocity of bob after collision,

v1 = 5 g R = 5 * 1 0 * 0 . 5 = 5 m / s

From Conservation of Momentum,

1 0 1 0 0 0 * v = 1 0 1 0 0 0 * 1 0 0 + 1 * 5  

->v = 400 m/s  

 

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

λ 4 = l λ = 4 l

f = v λ = v 4 l l = v 4 f
l = 3 4 4 * 2 5 = 0 . 3 4 m = 3 4 c m

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Potential Energy is maximum at extreme position

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

x 1 = 5 s i n ( 2 π t + π 4 )

x 2 = 5 2 ( s i n 2 π t + c o s 2 π t )

= 1 0 s i n ( 2 π t + π 4 )

x 2 , m a x = 2 . x 1 , m a x

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

L e t , E m a g b T c

So, [ E ] = [ m ] a [ g ] b [ T ] c  

M L 2 T 2 = M a ( L T 2 ) b T c  

Thus, E = k mg2 T2

Δ E E = 2 . Δ g g + 2 . Δ T T   

Percentage error in E = 2 * 4 + 2 * 3 = 14%

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