2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    The number of beats per second resulting from the vibration\[{{x}_{1}}=a\,\cos \,500\,\pi t\] and\[{{x}_{2}}=a\,\cos \,508\,\pi t\] is

    A)  zero                                     

    B)  2

    C)  4                            

    D)         8

    Correct Answer: C

    Solution :

    Given,   \[{{x}_{1}}=a\cos 500\pi t\]                 \[{{x}_{2}}=a\cos 508\pi t\]                 \[{{\omega }_{1}}=2\pi {{n}_{1}}=500\pi \]                 \[{{\omega }_{2}}=2\pi {{n}_{2}}=508\pi \]                 \[n=\frac{{{n}_{2}}-{{n}_{1}}}{2}=\frac{508-500}{2}\]                 \[n=\frac{8}{2}=4\]


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