Question

A body of mass 0.2 kg performs linear S.H.M.
It experiences a restoring force of 0.2 N, when
its displacement from the mean position is 4
cm. Determine (i) force constant (ii) period of
S.H.M and (iii) acceleration of the body when the
displacement from the mean position is 1 cm.
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Answer 1

Answer:

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Answer 2

Given :

Mass (m) = 0.2 Kg

Restoring Force (F) = 0.2 N

To Find :

(i) force constant

(ii) period of S.H.M and

(iii) acceleration of the body when the displacement from the mean position is 1 cm.

Solution :

(i) We know, Force constant (K) = $\frac{Force (F)}{Displacement (x)}$

As F = 0.2 N and x = 4 cm = 4 × 10⁻² m. Therefore,

  Force Constant (K) =  $\frac{0.2}{4 * 10^-^2}$

                                  =  5 N/m

Hence, the force constant is 5 N/m.

(ii) We know,

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

                                         = $\sqrt{\frac{5}{0.2} }$

                                         = 5 rad/sec

∴ Time Period = $\frac{2\pi }{w}$

                        =$\frac{2 * 3.14}{5}$

                        = 1.256 sec

Hence, Time period is 1.256 seconds.

(iii) We know, Acceleration = ω²x

where, ω = Angular frequency and x = Displacement

As, ω = 5 and x = 1 cm = 0.01 m

Acceleration = ω²x

                     =  (5)² × 0.01

                     =  0.25 m/sec

Therefore, acceleration of the body is 0.25 m/sec