Question #160919
A 200g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250s. The total energy of the system is 2.00J. Find
(a) the force constant of tge spring
(b) the amplitude of the motion.
1
Expert's answer
2021-02-11T10:43:19-0500

(a)

T=2πmk,T=2\pi\sqrt{\dfrac{m}{k}},k=4π2mT2,k=\dfrac{4\pi^2m}{T^2},k=4π20.2 kg(0.25 s)2=126.3 Nm.k=\dfrac{4\pi^2\cdot0.2\ kg}{(0.25\ s)^2}=126.3\ \dfrac{N}{m}.

(b)

E=12kA2,E=\dfrac{1}{2}kA^2,A=2Ek,A=\sqrt{\dfrac{2E}{k}},A=22.0 J126.3 Nm=0.177 m.A=\sqrt{\dfrac{2\cdot2.0\ J}{126.3\ \dfrac{N}{m}}}=0.177\ m.

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