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Part b) Name three examples of simple harmonic motion. State Hooke's Law and calculate the length of the spring when the applied force is 14 N. The natural length of the spring is 3.4 m and when a for

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3.456 m\text{3.456 m}

5.29×1010 Nm−25.29 \times 10^{10} \text{ Nm}^{-2}

Step-by-step solution

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b) Three examples of simple harmonic motion are:

  • The oscillation of a simple pendulum (for small angles).
  • The oscillation of a mass attached to a spring.
  • The vibration of a tuning fork.
  1. Hooke's Law states that within the elastic limit, tensile stress is directly proportional to tensile strain. Mathematically, this is expressed as:
σ=Eϵ\sigma = E \epsilon

where σ\sigma is the tensile stress, ϵ\epsilon is the tensile strain, and EE is Young's modulus (the constant of proportionality).

  1. Step 1: Calculate the spring constant (kk). Given: Natural length, L0=3.4 mL_0 = 3.4 \text{ m} Extension, ΔL1=0.02 m\Delta L_1 = 0.02 \text{ m} Force, F1=5 NF_1 = 5 \text{ N}

Using Hooke's Law, F=kΔLF = k \Delta L:

k=F1ΔL1k = \frac{F_1}{\Delta L_1} k=5 N0.02 mk = \frac{5 \text{ N}}{0.02 \text{ m}} k=250 N/mk = 250 \text{ N/m}

  1. Calculate the new extension (ΔL2\Delta L_2) when the applied force is 14 N14 \text{ N}.

    Given:
    New force, F2=14 NF_2 = 14 \text{ N}
    Spring constant, k=250 N/mk = 250 \text{ N/m}

    Using Hooke's Law:

    ΔL2=F2k\Delta L_2 = \frac{F_2}{k} ΔL2=14 N250 N/m\Delta L_2 = \frac{14 \text{ N}}{250 \text{ N/m}} ΔL2=0.056 m\Delta L_2 = 0.056 \text{ m}

  2. Calculate the new total length (L2L_2).

    L2=L0+ΔL2L_2 = L_0 + \Delta L_2 L2=3.4 m+0.056 mL_2 = 3.4 \text{ m} + 0.056 \text{ m} L2=3.456 mL_2 = 3.456 \text{ m}

    The length of the spring when the applied force is 14 N14 \text{ N} will be 3.456 m\boxed{\text{3.456 m}}.

    1. Step 1: Identify the given values. Tensile stress, σ=3.7×109 Nm−2\sigma = 3.7 \times 10^9 \text{ Nm}^{-2} Length increases by 7% of its original length. This means the tensile strain (ϵ\epsilon) is 7%.

  3. Convert the percentage increase in length to tensile strain.

    Tensile strain, ϵ=change in lengthoriginal length\epsilon = \frac{\text{change in length}}{\text{original length}}
    If the length increases by 7%, then ΔLL0=0.07\frac{\Delta L}{L_0} = 0.07.
    So, ϵ=0.07\epsilon = 0.07.

  4. Calculate Young's modulus (EE).

    Young's modulus is defined as the ratio of tensile stress to tensile strain:

    E=σϵE = \frac{\sigma}{\epsilon} E=3.7×109 Nm−20.07E = \frac{3.7 \times 10^9 \text{ Nm}^{-2}}{0.07} E≈5.2857×1010 Nm−2E \approx 5.2857 \times 10^{10} \text{ Nm}^{-2}

  5. Round the result to an appropriate number of significant figures.

    E≈5.29×1010 Nm−2E \approx 5.29 \times 10^{10} \text{ Nm}^{-2}

    The Young's modulus of the wire is 5.29×1010 Nm−2\boxed{5.29 \times 10^{10} \text{ Nm}^{-2}}.

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