A 5.00-kg mass attached to a horizontal spring oscillates back and forth in simp

Question

A 5.00-kg mass attached to a horizontal spring oscillates back and forth in simple harmonic motion with amplitude of 0.200 m. If the spring has a force constant of 75.0 N/m, calculate a) the number of oscillations the spring does in 1.00 second. b) total mechanical energy of the spring-mass system. The figure below shows a snapshot of a proton moving at velocity v = 200 m/s toward a long straight wire with current i = 350 mA. At the instant shown, the proton's distance from the wire is d = 2.89 cm. a) Find the direction and magnitude of the magnetic field created by the wire at the position of the proton. b) Find the direction and magnitude of the magnetic force of the magnetic field of the wire on the proton. The proton's charge is 1.60 times 10^-19 C.

Explanation / Answer

6.(a)

Number of oscillations / 1 s = frequency = 1 / T

Where T = period of oscillation

T = 2*pi*sqrt (m / k)

T = 2*3.14*sqrt (5 / 75) = 1.62 s

so, f = 1 / 1.62

f = 1.273 oscillations / s

(b)

Total machanical energy,

Et = (1/2)kA^2

Et = (1/2)*75*(0.20)^2

Et = 1.5 J

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