Question
The time period of a mass hanging from an ideal spring is 2 seconds. If along with it 2 kg if the mass is added and the time period becomes 3 seconds find the value of $m$.
✓
Answer
When a mass of $m$ hung then the time period is T , then
$T=2 \pi \sqrt{\frac{m}{k}}\ldots\ldots (1)$
Where, $k=$ force constant of the spring $(m+2) kg$ 's mass hung then time period is $T ^{\prime}$ so,
$T^{\prime}=2 \pi \sqrt{\frac{m+2}{k}}\ldots\ldots (2)$
Equation (2) divided by (1)
$\frac{T^{\prime}}{T}=\sqrt{\frac{m+2}{m}}$
Given: $T =2$ seconds, $T ^{\prime}=3$ seconds
$\frac{3}{2}=\sqrt{\frac{m+2}{m}}=\frac{9}{4}=\frac{m+2}{m}$
$\text {or}\quad 9 m=4 m+8$
$\Rightarrow5 m=8$
$\Rightarrow m=\frac{8}{5}=1.6 kg$
$T=2 \pi \sqrt{\frac{m}{k}}\ldots\ldots (1)$
Where, $k=$ force constant of the spring $(m+2) kg$ 's mass hung then time period is $T ^{\prime}$ so,
$T^{\prime}=2 \pi \sqrt{\frac{m+2}{k}}\ldots\ldots (2)$
Equation (2) divided by (1)
$\frac{T^{\prime}}{T}=\sqrt{\frac{m+2}{m}}$
Given: $T =2$ seconds, $T ^{\prime}=3$ seconds
$\frac{3}{2}=\sqrt{\frac{m+2}{m}}=\frac{9}{4}=\frac{m+2}{m}$
$\text {or}\quad 9 m=4 m+8$
$\Rightarrow5 m=8$
$\Rightarrow m=\frac{8}{5}=1.6 kg$
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