Time period of block springs SHM

  • Thread starter Jahnavi
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  • #1
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Homework Statement


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Homework Equations




The Attempt at a Solution



Length of the unstretched springs is L .Suppose block is pushed towards C by a small distance x . This causes a change in length of springs B and C . Their new length is L+∆L .Consider spring B. If the new length makes a small angle θ with the vertical .

I need to find relation between ∆L and x .

Applying cosine rule ,

x2 = (L+∆L)2 + L2 -2L(L+∆L)cosθ

For small angles , cosθ≈1.

Using this I get ∆L = x .

This is not correct . What is the mistake ?
 

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Answers and Replies

  • #2
haruspex
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For small angles , cosθ≈1.
Perhaps that is not exact enough here.
How about applying the cosine rule using the known angle?
 
  • #3
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Perhaps that is not exact enough here.

Could you explain what do you mean by this .

Should I use cosθ ≈ 1-θ2/2 , but then what value should I put for θ ?
 
  • #4
haruspex
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Should I use cosθ ≈ 1-θ2/2
Yes, but as you say the result is unhelpful, so why not try the alternative I suggested.
 
  • #5
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How about applying the cosine rule using the known angle?

That does give correct result :smile: Thank you .

This is one of the tougher problems in the book and the hint along with question is to use small angle approximation . This is why I used angle θ as it seemed to be the only small angle in the triangle formed .

If you can think of some why how to use small angle approximation , please let me know .
 
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  • #6
haruspex
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how to use small angle approximation
No, it does not seem useful here.
A simpler way is to drop a perpendicular from the original position of the mass to a stretched spring and observe that this produces a roughly 45-45-90 triangle with x as the hypotenuse and ΔL on each of the other sides.
 
  • #7
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A simpler way is to drop a perpendicular from the original position of the mass to a stretched spring and observe that this produces a roughly 45-45-90 triangle with x as the hypotenuse and ΔL on each of the other sides.

This is so good :smile:

How did you get this ? It is not obvious to me . How is length of the perpendicular ∆L (or how are the two angles 45° each ) ?
 
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  • #8
haruspex
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This is so good :smile:

How did you get this ? It is not obvious to me . How is length of the perpendicular ∆L (or how are the two angles 45° each ) ?
The perpendicular will meet the stretched spring roughly L from the spring's anchor point, leaving ΔL on the other side, from the perpendicular to the mass. Since that small triangle is nearly isosceles, the perpendicular itself is length ΔL.
 
  • #9
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The perpendicular will meet the stretched spring roughly L from the spring's anchor point,

How ?

Since that small triangle is nearly isosceles,

How ?
 
  • #10
haruspex
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The perpendicular will meet the stretched spring roughly L from the spring's anchor point
To be exact, L cos(θ) ≈ L.
Since that small triangle is nearly isosceles
Angles are 90, 45-θ, 45+θ.
 
  • #11
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To be exact, L cos(θ) ≈ L.

Angles are 90, 45-θ, 45+θ.

OK . Thanks :smile:
 

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