Question

At the distance from the mean position
is the kinetic energy of a partical performing S.H.M. of amplitude 8 cm , three times it's potential energy?​

Answer 1

Answer:

• The Distance from the mean position (x) is 4 cm.

Given:

1. Amplitude (A) = 8 cm.
2. Distance from mean position = x

Explanation:

$\rule{300}{1.5}$

From the formula we know,

⇒ K.E = 1 / 2 m ω² (A² - x²)

Where,

• m Denotes Mass of the body.
• ω Denotes Angular velocity.
• A Denotes Amplitude.
• x denotes Distance from mean position.

⇒ P.E = 1 / 2 m ω² x²

Where,

• m Denotes Mass of the body.
• ω Denotes Angular velocity.
• x denotes Distance from mean position.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

According to the Question,

⇒ K.E = 3 P.E

⇒ 1 / 2 m ω² (A² - x²) = 3 × (1 / 2 m ω² x²)

⇒ 1 / 2 m ω² (A² - x²) = 3 / 2 m ω² x²

Cancelling the common terms.

⇒ (A² - x²) = 3 x²

⇒ 3 x² + x² = A²

⇒ 4 x² = A²

⇒ x² = A² / 4

⇒ x = √ ( A² / 4 )

⇒ x = A / 2

Substituting the value of Amplitude,

⇒ x = 8 cm / 2

⇒ x = 8 / 2

⇒ x = 4

⇒ x = 4 cm

∴  The Distance from the mean position (x) is 4 cm.

$\rule{300}{1.5}$

Answer 2

$\huge\underline\mathtt\red{Answer:-}$

The Distance from the mean position (x) is 4 cm.