Question

the total enegry of a body performing SHM is E.then average kinetic energy of the body , over a period is​

Answer 1

Answer:

TotalenergyisSHME=

2

1

mw

2

a

2

(wherea=amplitude)

potentialenergyU=

2

1

mw

2

(a

2

−y

2

)

=E−

2

1

mw

2

y

2

Wheny=

2

a

U=E−

2

1

mw

2

(

4

a

2

)=E−

4

E

=

4

3E

Answer 2

Concept Introduction:-

The object is it's total energy it has due to the motion in physics is known as kinetic energy.

Given Information:-

We have been given that total enegry of a body performing SHM is E.

To Find:-

We have to find that average kinetic energy of the body, over a period is​

Solution:-

According to the problem

Total energy is $SHM E=\frac{1}{2}\m w^{2} a^{2}$ (where $a=$ amplitude) potential energy

$U=\frac{1}{2} m w^{2}\left(a^{2}-y^{2}\right)$

$=\mathrm{E}-\frac{1}{2} \mathrm{m} \mathrm{w}^{2} \mathrm{y}^{2}$

When $y=\frac{a}{2}$

$U=E-\frac{1}{2} m w^{2}\left(\frac{a^{2}}{4}\right)=E-\frac{E}{4}=\frac{3 E}{4}$.

Final Answer:-

The correct answer is $\frac{3E}{4}$.

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