Question 250: Oscillations

Question

A mass $x$ gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes $\frac{4 \mathrm{~T}}{3}$. The ratio of $\mathrm{Y} / \mathrm{x}$ is

Accepted Answer

Let the force constant (spring constant) be $k$.
Step 1: Write the formula for the time period of S.H.M.:
For a mass $m$ attached to a spring of force constant $k$,
$$ T = 2\pi \sqrt{\frac{m}{k}} $$
Given:
Initial mass $= x$ grams
New mass $= x + Y$ grams
Initial period $= T$
New period $= \frac{4T}{3}$
Step 2: Express both situations using the time period formula.
When mass is $x$ grams:
$$    T = 2\pi \sqrt{\frac{x}{k}}    $$
When mass is $x + Y$ grams:
$$    \frac{4T}{3} = 2\pi \sqrt{\frac{x + Y}{k}}    $$
Step 3: Divide the two equations.
$$ \frac{\frac{4T}{3}}{T} = \frac{2\pi \sqrt{\frac{x + Y}{k}}}{2\pi \sqrt{\frac{x}{k}}} $$
$$ \frac{4}{3} = \sqrt{\frac{x + Y}{x}} $$
Step 4: Square both sides to remove the square root.
$$ \left(\frac{4}{3}\right)^2 = \frac{x + Y}{x} $$
$$ \frac{16}{9} = \frac{x + Y}{x} $$
Step 5: Rearranging to find $Y/x$.
$$ x + Y = \frac{16}{9} x $$
$$ Y = \frac{16}{9} x - x $$
$$ Y = \left( \frac{16}{9} - 1 \right)x = \frac{7}{9} x $$
$$ \frac{Y}{x} = \frac{7}{9} $$
Step 6: Final answer
The required ratio is
$$ \boxed{7:9} $$
Correct option: A

Answer

$7: 9$

Answer

$5: 4$

Answer

$3: 2$

Answer

$8: 7$

Oscillations

A mass x gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes \frac{4 \mathrm{~T}}{3}. The ratio of \mathrm{Y} / \mathrm{x} is

Choose the correct answer: