Time period of block springs SHM

[Jahnavi]

Homework Statement

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 ?

 

[haruspex]

For small angles , cosθ≈1.

Perhaps that is not exact enough here.
How about applying the cosine rule using the known angle?

 

[Jahnavi]

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 ?

 

[haruspex]

Should I use cosθ ≈ 1-θ²/2

Yes, but as you say the result is unhelpful, so why not try the alternative I suggested.

 

[Jahnavi]

How about applying the cosine rule using the known angle?

That does give correct result 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 .

 

[haruspex]

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.

 

[Jahnavi]

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

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 ) ?

 

[haruspex]

This is so good

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.

 

[Jahnavi]

The perpendicular will meet the stretched spring roughly L from the spring's anchor point,

How ?

Since that small triangle is nearly isosceles,

How ?

 

[haruspex]

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.

 

[Jahnavi]

To be exact, L cos(θ) ≈ L.

Angles are 90, 45-θ, 45+θ.

OK . Thanks