Question

A mass of 100 g attached to a massless
spring of stiffness constant 10 N/m is under
damped motion. If the energy of the
oscillator decays to 1/e times of its initial
value in 50 s then the damping constant is​

Answer 1

Answer:

For damped oscillation,

(a)T=

ω

2π

T=

m

K

−(

2m

l

)

2

2π

T=

200

80×1000

−(

1000×2×200

40×1000

)

2

2×3.14

T=

400−

100

1

6.28

T=0.314s

(b)F×A

2

If amplitude gets reduced to half the force will reduce to one fourth.

F∝a

a∝ω

2

So ω will become half.

T=

ω

2π

T=2×0.314=0.628s

Explanation:

HERE IS YOUR ANSWER

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