Question

Amplitude of a particle in SHM is 6 cm. If
instantaneous potential energy is half the
total energy, then distance of particle
from its mean position is​

Answer 1

Answer:

The potential energy in SHM is given by ,

U=1/2ky^2

where , k= force constant

y= distance of particle from mean position

Total energy in SHM is given by ,

E=1/2ka^2

where, a= amplitude in SHM

given , U=E/2

1/2ky^2

=(1/2)(1/2)ka^2

or y^2=a^2 /2

now , given that a=6cm ,

Hence, y^2 =6^2/2=18

or y= √18

. =4.2cm

Answer 2

$\Large\bf{\color{indigo}GiVeN,}$

• Amplitude (a) of a particle in S.H.M is 6 cm.

• Instantaneous potential energy is ½ the total energy.

$\bf\blue{We\:know\:that,}$

❶ The potential energy in S.H.M is given by,

$\red\bigstar\:\:{\green{\boxed{\bf{\color{peru}U\:=\:\dfrac{1}{2}\:k\:y^2}}}}$

$\bf\pink{Where,}$

• k is the force constant.

• y is the distance of particle from mean position.

❷ Total energy in S.H.M is given by,

$\green\bigstar\:\:{\pink{\boxed{\bf{\color{olive}T.E\:=\:\dfrac{1}{2}\:k\:a^2}}}}$

$\bf\purple{Where,}$

• a is the amplitude in S.H.M.

$\bf\orange{According\:to\:the\:question,}$

$:\implies\:\:\bf{U\:=\:\dfrac{T.E}{2}\:}$

$:\implies\:\:\bf{\dfrac{1}{2}\:k\:y^2\:=\:\dfrac{1}{2}\times\Big(\dfrac{1}{2}\:k\:a^2\Big)}$

$:\implies\:\:\bf{k\:y^2\:=\:\dfrac{1}{2}\:k\:a^2}$

$:\implies\:\:\bf{y^2\:=\:\dfrac{1}{2}\times{6}^2}$

$:\implies\:\:\bf{y^2\:=\:\dfrac{1}{2}\times{36}}$

$:\implies\:\:\bf{y^2\:=\:18}$

$:\implies\:\:\bf{y\:=\:\sqrt{18}}$

$:\implies\:\:\bf{y\:=\:3\sqrt{2}}$

$:\implies\:\:\bf\green{y\:=\:4.24\:cm}$

$\Large\bold\therefore$ The distance of particle from its mean position is 4.24 cm.