P140 solutions Dec. 5

# P140 solutions Dec. 5 - SOLUTIONS – DECEMBER 5 15.7...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTIONS – DECEMBER 5 15.7. IDENTIFY: Use Eq.(15.1) to calculate v . T = 1/ f and k is defined by Eq.(15.5). The general form of the wave function is given by Eq.(15.8), which is the equation for the transverse displacement. SET UP: v = 8.00 m/s, A = 0.0700 m, λ = 0.320 m EXECUTE: (a) v = f λ so f = v / λ = (8.00 m/s)/(0.320 m) = 25.0 Hz T = 1/ f = 1/ 25.0 Hz = 0.0400 s k = 2 π / λ = 2 π rad/0.320 m = 19.6 rad/m (b) For a wave traveling in the − x-direction, y ( x , t ) = A cos2 π ( x / λ + t / T ) (Eq.(15.8).) At x = 0, y (0, t ) = A cos2 π ( t / T ), so y = A at t = 0. This equation describes the wave specified in the problem. Substitute in numerical values: y ( x , t ) = (0.0700 m)cos(2 π ( x / 0.320 m + t / 0.0400 s)). Or, y ( x , t ) = (0.0700 m)cos((19.6 m − 1 ) x + (157 rad/s) t ). (c) From part (b), y = (0.0700 m)cos(2 π ( x / 0.320 m + t / 0.0400 s)). Plug in and x = 0.360 m t = 0.150 s: y = (0.0700 m)cos(2 π (0.360 m/0.320 m + 0.150 s/0.0400 s)) y = (0.0700 m)cos[2 π (4.875 rad)](4....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

P140 solutions Dec. 5 - SOLUTIONS – DECEMBER 5 15.7...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online