Question

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Two springs fixed at one end are stretched by 5cm and 1cm, respectively, wher
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:
A) 1:2
B) 1:2
C) 2:1
D) 2:1​

Answer 1

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=mgkx=mg

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

x

mg

For first spring,

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

1

=

5×10

−2

0.5×9.8

k_{1}=98\ N/mk

1

=98 N/m

First second spring,

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

2

=

10×10

−2

1×9.8

k_{2}=98\ N/mk

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}}T=2π

k

m

For first spring,

T_{1}=2\pi\sqrt{\dfrac{0.5}{98}}T

1

=2π

98

0.5

....(I)

For second spring,

T_{2}=2\pi\sqrt{\dfrac{1}{98}}T

2

=2π

98

1

.....(II)

divided equation (I) by (II)

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

T

2

T

1

=

1×98

0.5×98

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

T

2

T

1

=

2

1

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