Question

2. Two springs fixed at one end are stretched by 5 cm and 10 cm,
respectively, when masses 0.5 kg and 1 kg are suspended at their
lower ends. When displaced slightly from their mean positions and
released, they will oscillate with time periods in the ratio:​

Answer 1

Answer:

$1 \div \sqrt{2}$

Explanation:

$time \: propotional \: to \: \sqrt{(l \div g)}$

t1/t2 = square root(5/10)

Answer 2

Answer:

The ratio of the time period is 1:√2.

Explanation:

Given that,

First mass = 0.5 kg

Second mass = 1 kg

First stretch = 5 cm

Second stretch = 10 cm

We need to calculate the spring constant

Using relation of restoring and newton's second law

$kx=mg$

$k=\dfrac{mg}{x}$

For first spring,

$k_{1}=\dfrac{0.5\times9.8}{5\times10^{-2}}$

$k_{1}=98\ N/m$

First second spring,

$k_{2}=\dfrac{1\times9.8}{10\times10^{-2}}$

$k_{2}=98\ N/m$

We need to calculate the ratio of time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

For first spring,

$T_{1}=2\pi\sqrt{\dfrac{0.5}{98}}$....(I)

For second spring,

$T_{2}=2\pi\sqrt{\dfrac{1}{98}}$.....(II)

divided equation (I) by (II)

$\dfrac{T_{1}}{T_{2}}=\sqrt{\dfrac{0.5\times98}{1\times98}}$

$\dfrac{T_{1}}{T_{2}}=\dfrac{1}{\sqrt{2}}$

Hence, The ratio of the time period is 1:√2.