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A mass m is suspended from the two coupl...

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are `k_(1) "and" k_(2).` The time period of the suspended mass will be

A

`T= 2pi sqrt(((m)/(K_1- K_2)))`

B

`T= 2pi sqrt(((m)/(K_1+K_2)))`

C

`T= 2pi sqrt(((m(K_1 + K_2))/(K_1 K_2)))`

D

`T= 2pi sqrt(((m K_1 K_2)/(K_1 + K_2)))`

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • A mass m is suspended from the two coupled springs connected in series. The force constant for spring are k_(1) and k_(2) . The time period of the suspended mass will be:

    A
    `T=2pisqrt((m)/(k_(1)-k_(2)))`
    B
    `T=2pisqrt((mk_(1)k_(2))/(k_(1)+k_(2)))`
    C
    `T=2pisqrt((m)/(k_(1)+k_(2)))`
    D
    `T=2pisqrt((m(k_(1)+k_(2)))/(k_(1)k_(2)))`

  • A mass m is suspended by means of two coiled springs which have the same length in unstretched condition as shown in figure. Their force constants are k_(1) and k_(2) respectively. When set into vertical vibrations, the period will be

    A
    `2pisqrt(((m)/(k_(1)k_(2))))`
    B
    `2pisqrt(m((k_(1))/(k_(2))))`
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    D
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  • A mass m is suspended separately by two different springs of spring constant k_(1) and k_(2) given the time period t_(1) and t_(2) respectively. If the same mass m is shown in the figure then time period t is given by the relation

    A
    `t = t_(1) + t_(2)`
    B
    `t =( t_(1)t_(2))/(t_(1) + t_(2))`
    C
    `t^(2) = t_(1)^(2) + t_(2)^(2)`
    D
    `t^(-2) = t_(1)^(-2) + t_(2)^(-2)`

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