BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    Two light springs of force constants \[{{k}_{1}}\] and \[{{k}_{2}}\] and a block of mass m are in one line AS on a smooth horizontal table, such that one end of each spring is fixed to rigid support and other end is attached to block of mass m kg as shown in figure. The frequency of vibration is:

    A) \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}\]

    B) \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{m}}\]

    C) \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}-{{k}_{2}}}{m}}\] 

    D) none of these

    Correct Answer: A

    Solution :

    When the oscillating mass m is at a distance \[x\] towards right from its equilibrium position, then the spring \[{{k}_{1}}\]is stretched through \[x,\]while \[{{k}_{2}}\]is compressed through \[x.\]Hence, restoring force exerted by each spring on the mass m is in the same direction tending to bring it in its equilibrium position. Let \[{{F}_{1}}\]and \[{{F}_{2}}\]be the restoring forces produced in the springs of force constants \[{{k}_{1}}\]and\[{{k}_{2}},\] then \[{{F}_{1}}=-{{k}_{1}}x\]and \[{{F}_{2}}=-{{k}_{2}}x\] Total restoring force acting on the mass is \[F={{F}_{1}}+{{F}_{2}}=-{{k}_{1}}x-{{k}_{2}}x=-({{k}_{1}}+{{k}_{2}})x=-kx\] \[\therefore \]  \[k={{k}_{1}}+{{k}_{2}}\] Hence, time period is                                   \[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{m}{{{k}_{1}}+{{k}_{2}}}}\]                     Also, frequency                                 \[n=\frac{1}{T}.=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}+{{k}_{2}}}{m}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner