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 t1 will it first reach the limit of its motion to the right?

Answer 1

Answer:

t = 0.325343 sec

Explanation:

given

mass of block(m) = 500 g

force constant  (k) = 25.0 N/m

we know simple harmonic motion equation as

y = A sin (ωt + Φ)

 where  Φ = $sin^{-1} (-4/6)$

so we get

Φ = - 0.729727656 rad  

and we know

$\omega = \sqrt{\frac{k}{m}$

$y = A sin (t\sqrt\frac{k}{m} - 0.729727656)$

now substitute the values we get as

y = 0.06 m

$0.06 = 0.06 sin (t\sqrt(\frac{25}{0.5}) - 0.729727656)$

$(t\frac{25}{0.5} - 0.729727656) = \frac{\pi}{2}$

on solving these equation we get

t = 0.325343 sec