When a particular mass-spring system is started by stretching the spring 3 cm, one complete oscillation takes 4 seconds. (a) If

Question

When a particular mass-spring system is started by stretching the spring 3 cm, one complete oscillation takes 4 seconds.

(a) If instead the initial stretch had been 5 cm, how long would it take for one complete oscillation?
Time for one oscillation = s

(b) If the mass were doubled, what would the period of the system be?
Period = s

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Serenity 31 mins 2021-12-27T06:56:05+00:00 1 Answer 0 views 0

Answers ( )

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    2021-12-27T06:57:16+00:00

    Answer:

    A. 5.16 s.

    B. 5.66 s.

    Step-by-step explanation:

    A.

    For a simple harmonic motion,

    T = 2pi (sqrt * (l/g))

    Given:

    L1 = 3 cm

    T1 = 4 s

    L2 = 5 cm

    T2 = ?

    4 = 2pi*sqrt(3/g)

    g = 7.4

    At, L2,

    T2 = 2pi*sqrt(5/7.4)

    = 5.16 s.

    B.

    M1 = M1

    M2 = 2*M1

    For a simple harmonic motion,

    T = 2pi (sqrt * (m/k))

    4 = 2pi (sqrt * (M1/k))

    M1/k = 0.405

    Inputting the above values,

    T2 = 2pi (sqrt * (2*M1/k))

    = 2pi (sqrt * (2 * 0.405))

    = 5.66 s.

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45:7+7-4:2-5:5*4+35:2 =? ( )