A 50.0-g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m. Its initial displacement is 0.300 m. A damping force

Question

A 50.0-g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m. Its initial displacement is 0.300 m. A damping force Fx=−bvxF

x

​

=−bv

x

​

acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant b.

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Yến Oanh 4 years 2021-08-25T10:33:05+00:00 1 Answers 8 views 0

Answers ( )

  1. ngockhue
    0
    2021-08-25T10:34:19+00:00
    Reply

    Answer:

    The correct answer will be “0.022 kg/s“.

    Explanation:

    The given values are:

    Initial amplitude,

    A₁ = 0.3 m

    Final amplitude,

    A₂ = 0.1 m

    Mass of a egg,

    m = 50.0 g i.e., 0.050 kg

    Time,

    t = 5 sec

    As we know,

    ⇒  A_{2}=A_{1}e^{(-\frac{b}{2m})t}

    So that, the damping constant will be:

    ⇒  b=\frac{2m}{t} log \frac{A_{1}}{A_{2}}

    On putting the estimated values, we get

    ⇒  b=\frac{2\times 0.050}{5}log \frac{0.3}{0.1}

    ⇒     =\frac{0.1}{5} log \frac{0.3}{0.1}

    ⇒     =0.02 \ log 3

    ⇒     =0.0219 \ kg/sec \ or \ 0.022 \ kg/s

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