Question

A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass when the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3\,s$. When the mass $m$ is increased by $1\,kg$, the time period of oscillations becomes $5\, s$. The value of $m$ in $kg$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{16}{9}$
• D. $\frac{9}{16}$

Answer 1

The Correct Option is D

Solving the Equation for \( m \) 

We are given two equations involving \( m \) and \( k \), and we need to solve for \( m \).

Step 1: Given Equations

The first equation is:

\(3 = 2\pi \sqrt{\frac{m}{k}}\)

The second equation is:

\(5 = 2\pi \sqrt{\frac{m+1}{k}}\)

Step 2: Taking the Ratio of the Equations

Now, we take the ratio of the two equations to eliminate \( k \). This gives:

\(\frac{3}{5} = \sqrt{\frac{m}{m+1}}\)

Step 3: Squaring Both Sides

Next, we square both sides of the equation to remove the square root:

\(\left(\frac{3}{5}\right)^2 = \frac{m}{m+1}\)

This simplifies to:

\(\frac{9}{25} = \frac{m}{m+1}\)

Step 4: Solving for \( m \)

To solve for \( m \), we cross-multiply:

\(9(m+1) = 25m\)

Now expand the equation:

\(9m + 9 = 25m\)

Now, subtract \( 9m \) from both sides:

\(9 = 16m\)

Finally, solve for \( m \):

\(m = \frac{9}{16}\)

Conclusion:

The value of \( m \) is \( \frac{9}{16} \).