Question

A particle performing shm with amplitude 10 cm what distance from mean position the kinetic energy of the particle is size of its potential energy

Answer 1

The distance is x = ±5 cm

Explanation:

According to question :

1 / 2 mv^2 = 3 × 1 / 2 mω^2 x^2

⇒1 / 2 mω^2(A^2 − x^2)

= 3 / 2 m ω^2.x^2

A^2 = 4.x^2

x = ±A^2 ± 10 cm / 2

x = ±5 cm

Thus the distance is x = ±5 cm

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Answer 2

± 7 cm from mean position the kinetic energy of the particle is equal to its potential energy.

Explanation:

Given amplitude, A =10 cm.

The kinetic energy in SHM is given as

$K=\frac{1}{2}m\omega^2(A^2-x^2)$

Here, x is the distance from mean position.

The potential energy in SHM is given as

$U=\frac{1}{2}m\omega^2x^2$

As per question,

K = U

$\frac{1}{2}m\omega^2(A^2-x^2)=\frac{1}{2}m\omega^2x^2$

$A^2=2x^2$

substitute the values, we get

$x^2=\frac{(10cm)^2}{2}$

x = ± 7 cm

Thus, ± 7 cm from mean position the kinetic energy of the particle is equal to its potential energy.

#Learn More: Energy in SHM.