A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligibl

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 $3	ext{ s}$. When the mass $m$ is increased by $1	ext{ kg}$, the time period of oscillations becomes $5	ext{ s}$. The value of $m$ in kg is:

Accepted Answer

The time period ($T$) of a spring-mass system undergoing simple harmonic motion is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant.

1.  **Case 1:** For mass $m$, the time period is $3	ext{ s}$.
    $$ 3 = 2\pi \sqrt{\frac{m}{k}} $$

2.  **Case 2:** When the mass is increased by $1	ext{ kg}$, the new mass is $(m + 1)$ and the time period becomes $5	ext{ s}$.
    $$ 5 = 2\pi \sqrt{\frac{m+1}{k}} $$

3.  **Divide the two equations:**
    $$ \frac{3}{5} = \frac{2\pi \sqrt{m/k}}{2\pi \sqrt{(m+1)/k}} $$
    $$ \frac{3}{5} = \sqrt{\frac{m}{m+1}} $$

4.  **Solve for $m$:**
    Squaring both sides:
    $$ \frac{9}{25} = \frac{m}{m+1} $$
    $$ 9(m + 1) = 25m $$
    $$ 9m + 9 = 25m $$
    $$ 16m = 9 $$
    $$ m = \frac{9}{16} 	ext{ kg} $$

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