A particle oscillating simple harmonically with an amplitude 1.5cm
Answers
The particle will be acted upon at a displacement of 2.25 cm
Step-by-step explanation:
Correct statement:
A particle oscillating harmonically with an amplitude of 1.5 cm, has a maximum energy of 0.25 uj. At what displacement from the equilibrium position will the particle be acted upon by a force of 2.5 x 10-N?
Solution:
Given data:
A = 1.5 cm
Maximum energy E = 0.25 x 10^-6 J
E = 1 / 2 kA^2
E = 1 / 2 k 1.5 / 100 × 1.5 / 100
⇒0.25 × 10^−6 = 225 / 10^6 (1 / 2 k)
(1 / 2k) = 0.25 × 1 / 225 --------(1)
Now Force at x=x is given by
F = 1 / 2k x
⇒2.5 × 10^−5 = 0.25 × 1 / 225 x
⇒x = 225 × 10^−4 m =2.25 cm
Thus the particle will be acted upon at a displacement of 2.25 cm
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