Question

A block with a mass of 0.600 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilibrium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 35.0 cm from equilibrium is at t = 0.200 s.(a) What is the block's period of oscillation?(b) What is the the value of the spring constant?(c) What is the block's velocity at t = 0.200 s? (Indicate the direction with the sign of your answer.(d) What is the block's acceleration at t = 0.200 s? (Indicate the direction with the sign of your answer.

Answer 1

Answer:

Explanation:

The amplitude of the oscillation under SHM will be .5 m and the equation of

SHM can be written as follows

x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.

x = .5 cosωt

given , when t = .2 s , x = .35 m

.35 = .5 cos ωt

ωt = .79

ω = .79 / .20

= 3.95 rad /s

period of oscillation

T = 2π / ω

= 2 x 3.14 / 3.95

= 1.6 s

b )

ω = $\sqrt{\frac{k}{m} }$

ω² = k / m

k = ω² x m

= 3.95² x .6

= 9.36 N/s

c )

v = ω$\sqrt{(a^2-x^2)}$

At t = .2 , x = .35

v = 3.95 $\sqrt{.5^2-.35^2}$

= 3.95 x .357

= 1.41 m/ s

d )

Acceleration at x

a = ω² x

= 3.95 x .35

= 1.3825 m s⁻²