Physics

Question

What is the mass of the ball that was hung to the spring if the period is 4 sec and the spring constant is 80 N/m.

1 Answer

  • Answer:

    Approximately 32 kg

    Explanation:

    What we know are period (T) = 4 seconds and spring constant (k) = 80 N/m (Newton/meter)

    Therefore, from the formula of:

    [tex] \displaystyle{T = 2\pi \sqrt{ \dfrac{m}{k}} }[/tex]

    Substitute in known values:

    [tex] \displaystyle{4 \: \text{seconds} = 2\pi \sqrt{ \dfrac{m}{80 \: \text{N/m}} }}[/tex]

    Divide both sides by 2π:

    [tex] \displaystyle{ \dfrac{4 \: \text{seconds} }{2\pi}=\sqrt{ \dfrac{m}{80 \: \text{N/m}} }} \\ \\ \displaystyle{ \dfrac{2 \: \text{seconds} }{\pi}=\sqrt{ \dfrac{m}{80 \: \text{N/m}} }}[/tex]

    Square both sides:

    [tex] \displaystyle{ \left( \dfrac{2 \: \text{seconds} }{\pi} \right)^{2} = \left(\sqrt{ \dfrac{m}{80 \: \text{N/m}} } \right)^{2} } \\ \\ \displaystyle{ \dfrac{4 \: \text{s}^2 }{ {\pi}^{2} } = \dfrac{m}{80 \: \text{N/m}}}[/tex]

    Then move 80 N/m to multiply left side:

    [tex] \displaystyle{ \dfrac{4 \: \text{s}^2 }{ {\pi}^{2} } \cdot 80 \: \text{N/m}= m} \\ \\ \displaystyle{ \dfrac{320 }{ {\pi}^{2} } \: \text{kg}= m}[/tex]

    Since you didn't specify which digit of π to use so I will go with what most people use, π ≈ 3.14

    [tex] \displaystyle{ \dfrac{320}{ {(3.14)}^{2} } \: \text{kg} = m}[/tex]

    3.14*3.14 = 9.8596 but rounded to 9.86 through significant figure rule:

    [tex] \displaystyle{ \dfrac{320}{9.86} \: \text{kg} = m}[/tex]

    320 can be converted to scientific notation which is 3.2*10². Therefore, this has 2 figures. Hence, the result must be in 2 digits after division:

    [tex] \displaystyle{32 \approx m}[/tex]

    Therefore, the mass is around 32 kg.

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