JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period

    A) \[T=2\pi \sqrt{\frac{mg}{x(M+m)}}\]     

    B) \[T=2\pi \sqrt{\frac{(M+m)x}{mg}}\]

    C) \[T=\frac{\pi }{2}\sqrt{\frac{mg}{x(M+m)}}\]     

    D) \[T=2\pi \sqrt{\frac{M+m}{mg}}\]

    Correct Answer: B

    Solution :

    Let elongation in spring is \[{{x}_{0}}\] when M is suspended, then \[Mg=k{{x}_{0}}\] \[(M+m)g=K({{x}_{0}}+x)\Rightarrow kx=mg\] Time period of combined mass,\[T=2\pi \sqrt{\frac{M+m}{K}}\]\[\Rightarrow \]\[T=2\pi \sqrt{\frac{(M+m)x}{mg}}\]

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