Question

The total energy of a particle of mass 200 g, performing S.H.M. is 10−2 J. Find its maximum velocity and period if the amplitude is 7 cm

Answer 1

Explanation:

Hope this answer helps!

Given: E = 10-² J = 0.01 J

m = 7 cm = 0.07 m

a = 200 g = 0.2 kg

Solution:

K.E = 1/2mv²

0.01 = 1/2×0.2×v²

v² = (2×0.01)÷0.2

Therefore,

v² = 0.1 m/s

Also, v = ω × amplitude

0.01 = ω × 0.07

ω = 0.01÷0.07

ω = 1.4 rad/s

We have, T= 2π/ω

= (2×3.14)÷1.4

T = 4.4 sec

Answer 2

Given :

Mass (M) = 200 g = 0.2 Kg

Total Energy = 10⁻² J = 0.01 J

Amplitude (A) = 7 cm = 0.07 m

To Find :

(i) Maximum velocity

(ii) Time period

Solution :

We know,

   Total Energy (TE) = $\frac{1}{2} KA^2$

Where, K = Force constant; A = Amplitude

⇒ Total Energy (TE) =  $\frac{1}{2}$ × K × (0.07)²

⇒      0.01                  =  $\frac{1}{2}$ × K × 0.0049

⇒      K                      =  $\frac{0.02}{0.0049}$

∴       K                      =  4.08 N/m

Angular frequency (ω) = $\sqrt{\frac{K}{m} }$

                                     =  $\sqrt{\frac{4.08}{0.2} }$

                                     = 4.52 rad/sec

(i) Maximum velocity ($V_m_a_x$) = ωA

                                             =  4.52 × 0.07

                                             =  0.3164 m/sec

∴ Maximum velocity is 0.3164 m/sec

(ii) Time period (T) = $\frac{2\pi }{w}$

                              = $\frac{2 * 3.14}{4.52}$

                              = 1.39 seconds

∴ Time period of the given SHM is 1.39 seconds.