Question

The total energy of a body of mass 2 kg performing S.H.M. is 50 J . Find its speed while crossing the center of the path.

Answer 1

Answer:

Ans: 6.32 cm/s .

Hope it helps .

Explanation:

Answer 2

$\dag \: \underline{\sf AnsWer :}$

• We are given that, a body is performing a S.H.M whose total energy (TE) is 50 J and mass (m) is 2kg. We need to find its speed while crossing the center of the path. So, let's do it :

$: \implies \sf Total \: energy = \dfrac{1}{2} m \omega A^2$

$: \implies \sf Total \: energy = \dfrac{1}{2} m (A\omega)^{2}$

$: \implies \sf Total \: energy = \dfrac{1}{2} m (v)^{2} \qquad \bigg\lgroup\bf \because V_{max} = A \omega \bigg \rgroup$

$: \implies \sf 40= \dfrac{1}{2} \times 2 \times (v)^{2}$

$: \implies \sf 40 \times 2= 2 \times (v)^{2}$

$: \implies \sf 80 = 2 (v)^{2}$

$: \implies \sf v^{2} = \dfrac{80}{2}$

$: \implies \sf v^{2} = 40$

$: \implies \sf v= \sqrt{40}$

$: \implies \sf v= \sqrt{4 \times 10}$

$: \implies \sf v= \sqrt{2 \times 2 \times 10}$

$: \implies \sf v= 2 \sqrt{10}$

$: \implies \sf v= 2 \times 3.16$

$: \implies \underline{ \boxed{ \sf v= 6.32 \: m/s}}$