A Uniform Spring Of Length L Has Force Constant K. The Spring Is Cut Into Two Pieces Of Unstressed Lengths L1, L2 Where (L1/l2)= (N1/n2). Express The Force Constants K1 And K2 Of The Two Pieces In Terms Of K, N1 And N2

Q) A Uniform Spring Of Length L Has Force Constant K. The Spring Is Cut Into Two Pieces Of Unstressed Lengths L1, L2 Where (L1/l2)= (N1/n2). Express The Force Constants K1 And K2 Of The Two Pieces In Terms Of K, N1 And N2.

Ans)1) Relate spring lengths and force constant:

• The force constant reflects the stiffness of a spring and is inversely proportional to its length. This means:
  • K ∝ 1 / L

2. Consider the total spring constant:

• Since the original spring (length L) has force constant K, the total stored energy within it is KE = ½ K L².

3. Distribute energy after cutting:

• When the spring is cut into two pieces with lengths L1 and L2, the total stored energy remains the same. Therefore:
  • ½ K L² = ½ K₁ L₁² + ½ K₂ L₂²

4. Substitute length ratio:

• We are given L₁/L₂ = N₁/N₂. Square both sides and rearrange:
  • L₁² = (N₁²/N₂²) L₂²

5. Substitute and combine equations:

• Substitute L₁² from step 4 into the energy equation:

  • ½ K L² = ½ K₁ (N₁²/N₂²) L₂² + ½ K₂ L₂²

• Rearrange and simplify:

  • K₁ = K * (N₁² / N₂²) + K₂

6. Express K₂ in terms of K₁:

• Substitute K₂ from the simplified equation back into the initial energy equation:

  • ½ K L² = ½ K₁ (N₁² / N₂²) L₂² + ½ [K₁ - K * (N₁² / N₂²)] L₂²

• Solve for K₂:

  • K₂ = K * (N₂² / N₁²)

Therefore, the force constants of the two pieces are:

• K₁ = K * (N₁² / N₂² + 1)
• K₂ = K * (N₂² / N₁²)

These equations express K1 and K2 in terms of the original force constant K, the ratios of natural frequencies N1/N2, and the inherent relationship between force constant and spring length. Remember that N1 and N2 must be positive integers due to their physical meaning as natural frequencies.