A block of mass 500 g is attached to a horizontal spring, whose force constant i

Question

A block of mass 500 g is attached to a horizontal spring, whose force constant is 25.0 N/m . The block is undergoing simple harmonic motion with an amplitude of 6.00 cm . At t=0 the block is 4.00 cm to the left of its equilibrium position and is moving to the right. At what time t 1 will it first reach the limit of its motion to the right?

Part A

Which of the following quantities are known in the situation described in the problem introduction?

Check all that apply.

Check all that apply.

x0 , initial displacement t1 , time at which block reaches its far right position , phase angle m , mass of the block T , period of oscillation k , force constant of the spring A , amplitude of oscillation , angular frequency

Explanation / Answer

y = A sin (t + )

= arc sin (-4/6)

= - 0.729727656 rad

= (k/m)

y = A sin (t(k/m) - 0.729727656)

the time that it first reach the limit of its motion to the right if:

y = 0.06 m

0.06 = 0.06 sin (t(25/0.5) - 0.729727656)

(t(25/0.5) - 0.729727656) = ½

t = 0.325343 sec

t measured from the first time of mass in equilibrium system


period

T = 2/

T = 2/(k/m)

T = 2(m/k)

T = 2(0.5/25)

T = 0.8886 sec

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