What is the phase of the SHM at that point

[Voltrical]

Homework Statement

A mass is attached to a spring on a frictionless horizontal surface. The mass is pulled to stretch the spring by 0.1 m, & then gently released. A short time later, as the mass passes through the equilibrium position, its speed is 1 m/s.

Part a)

What is the speed of the mass at the point where the spring is compressed by 0.05 m?

Part b)

What is the phase of the SHM at that point, relative to a phase of zero at the time of release?

Homework Equations

The Attempt at a Solution

Part a)

A = 0.1 m
ωA =1 m/s
ω = 1 / 0.1 = 10 rad/s
y = 0.05 m
V = ω √ (A² - y²)
= 10 * [ √ ( 0.01 - 0.0025 ) ]
= 0.866 m/s, this answer is correct because my teacher gave us the answer.

Part b)

Answer = 120°, how to get this answer?

 

[tiny-tim]

Hi Voltrical!

Part b)

Answer = 120°, how to get this answer?

shm is amplitude times cosωt …

so won't it have something to do with cos120° ? ​