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A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is

  • Option 1)

    3/5

  • Option 2)

    25/9

  • Option 3)

    16/9

  • Option 4)

    5/3

 

Answers (1)

As we learnt in

Time period of oscillation for spring mass system -

T= 2\pi \sqrt{\frac{m}{K}}

- wherein

m = mass of block

K = spring constant

 

 

 

Initially,\; T=2\pi \sqrt{M/k}

Finally,\; \frac{5T}{3}=2\pi \sqrt{\frac{M+m}{k}}

\therefore \; \; \; \frac{5}{3}\times 2\pi \sqrt{\frac{M}{k}}=2\pi \sqrt{\frac{M+m}{k}}

or\; \; \; \frac{25}{9}\frac{M}{k}=\frac{M+m}{k}

or\; \; \; 9\, m+9\, M=25\, M

or\; \; \; \frac{m}{M}=\frac{16}{9}

Correct option is 3.


Option 1)

3/5

Option 2)

25/9

Option 3)

16/9

Option 4)

5/3

Posted by

Vakul

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