Question

The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path.​

Answer 1

Answer:

Given that an object in simple harmonic motion of mass 2 kg has a reserved energy of 40 j.

To find speed, we use the kinetic energy formula:

$\to \sf Kinetic \ energy = \dfrac{1}{2} \cdot mass \cdot speed^2$

Substitute.

$\to \sf 40J = \dfrac{1}{ \not{2} } \cdot \not{2} \cdot speed^2$

$\to \sf \sqrt{40} = speed$

$\to \sf 2\sqrt{10} \ m/s= speed$

The speed is 2√(10) m/s or 6.324 m/s.

Answer 2

Total energy of a body, E = 40 J

Mass of the body, m = 2 kg

Speed, $\sf v_{max}$ = ωA = ?

Now, T.E. = $\dfrac{1}{2}$ (m)(ωA)²

• T.E. = $\dfrac{1}{2}$(m)($\sf v_{max}$)²

• 40 = $\dfrac{1}{2}$(2)($\sf v_{max}$)²

• $\sf v_{max}$² = $\dfrac{40 × 2}{2}$

• $\sf v_{max}$ = √40

• $\sf v_{max}$ = 6.32 m/s

$\bold{Hope\;it \; helps\;!}$