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When a mass m is connected individually to two springs S1 and S2, the oscillation frequencies are ν1 and ν2 . If the same mass is attached to the two springs as shown in Fig. 14.3, the oscillation frequency would be

(a) ν1 + ν2

(b) v 1 2 + v 2 2

(c) ( 1 v 1 + 1 v 2 ) -1 

(d) v 1 2 - v 2 2

0 3 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    This is a multiple choice answer as classified in NCERT Exemplar

    (b) Keq= K1+K2

     

    Time period of oscillation of the spring block system

    T=2 π m K e q = 2 π m K 1 + K 2

    Frequency = 1/time = 1 2 π K 1 + K 2 m = equivalent oscillation frequency

    When the mass is connected to the springs individually

    v 1 = 1 2 π k 1 m
    v 2 = 1 2 π k 2 m

    From above equation

    v = 1 2 π k 1 m + k 2 m 1 2

    v = 1 2 π 4 π 2 v 1 2 1 + 4 π 2 v 2 2 1 1 2

    So v = v 1 2 + v 2 2

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