# Phase of the following Waves

## Homework Statement

See figure attached for problem statement.

## The Attempt at a Solution

How does one go about solving a problem like this? I don't see how they get actual values from just the photo.

Can someone explain?

#### Attachments

• MQ14.JPG
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collinsmark
Homework Helper
Gold Member
How does one go about solving a problem like this? I don't see how they get actual values from just the photo.

Can someone explain?
Look at the y-intercept. Each wave crosses the y-axis exactly halfway between the origin and the wave's maximum amplitude. What sort of "angle" causes the sine function to be half its maximum amplitude? Look at the y-intercept. Each wave crosses the y-axis exactly halfway between the origin and the wave's maximum amplitude. What sort of "angle" causes the sine function to be half its maximum amplitude? I know the angle you're talking about is 30 degrees.

When I look at it, it looks like the first wave has a phase of -45 degrees and the second a phase of -135 degrees, but those options aren't present...

Because at 45 degrees it would be half way to reaching its maximum value at 90 degrees.

SammyS
Staff Emeritus
Homework Helper
Gold Member
I know the angle you're talking about is 30 degrees.

When I look at it, it looks like the first wave has a phase of -45 degrees and the second a phase of -135 degrees, but those options aren't present...

Because at 45 degrees it would be half way to reaching its maximum value at 90 degrees.
Look at the distance (on the x-axis) from the x-intercept to the x value of the maximum. Then compare that and where the y-intercept is (x=0). Looks to me like x=0 is about 1/3 the distance from the x-intercept to x of the max. 45° is halfway from the x-intercept to where Sine has it's max. The value of sine(45°) = 1/√(2) ≈ 0.707. The y-intercept does not look (to me) like it's close to being 70% of the max value.

If collinsmark is talking about 30°, then he knows what he's talking about! Where else is sin(θ) = 1/2 ?