# hw12 - cerutti(cpc566 HW12 TSOI(58160 This print-out should...

This preview shows pages 1–3. Sign up to view the full content.

cerutti (cpc566) – HW12 – TSOI – (58160) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Radio waves travel at the speed of light: 3 × 10 5 km / s. What is the wavelength of radio waves re- ceived at 97 . 9 MHz on your FM radio dial? Correct answer: 3 . 06435 m. Explanation: Let : v = 3 × 10 5 km / s = 300 million m / s and f = 97 . 9 MHz = 97 . 9 million Hz . λ = v f = 300 million m / s 97 . 9 million Hz = 3 . 06435 m . 002 10.0 points Earthquakes produce two kinds of seismic waves: he longitudinal primary waves (called P waves) and the transverse secondary waves (called S waves). Both S waves and P waves travel through Earth’s crust and mantle, but at different speeds; the P waves are always faster than the S waves, but their exact speeds depend on depth and location. For the pur- pose of this exercise, we assume the P wave’s speed to be 9990 m / s while the S waves travel at a slower speed of 5980 m / s. If a seismic station detects a P wave and then 21 . 1 s later detects an S wave, how far away is the earthquake center? Correct answer: 314 . 344 km. Explanation: Let : v P = 9990 m / s , v S = 5980 m / s , and Δ t = 21 . 1 s . Suppose the earthquake happens at time t = 0 at some distance d . The P wave and the S wave are both emitted at the same time t = 0, but they arrive at different times, re- spectively t P = d v P and t S = d v S . The S wave is slower, so it arrives later than the P wave, the time difference being Δ t = d v S - d v P = d ( v P - v S ) v P v S . Consequently, given this time difference and the two waves’ speeds v P and v S , the earth- quake center is d = v P v S Δ t v P - v S = (9990 m / s) (5980 m / s) (21 . 1 s) 9990 m / s - 5980 m / s · 1 km 1000 m = 314 . 344 km away from the seismic station. 003 (part 1 of 2) 10.0 points A wave moving along the x axis is described by y ( x, t ) = A e ( x + vt ) 2 /b where x is in meters, t is in seconds, A = 4 m, v = 5 m / s, and b = 4 m 2 . Determine the speed of the wave. Correct answer: 5 m / s. Explanation: y ( x, t ) = A e ( x + vt ) 2 /b is of the form y ( x, t ) = f ( x + v t ) so it describes a wave at v = 5 m / s . 004 (part 2 of 2) 10.0 points Determine the direction of the wave. 1. - x direction correct 2. + x direction 3. - y direction

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

View Full Document
cerutti (cpc566) – HW12 – TSOI – (58160) 2 4. + y direction Explanation: f ( x + v t ) = f ( x - ( - v ) t ) so the wave is moving at - 5 m / s (in the negative direction). 005 (part 1 of 2) 10.0 points The time interval indicated on this diagram is G . + S 0 - S 0 S t (seconds) -→ G Which formula corresponds best to the di- agram? 1. S ( t ) = S 0 sin parenleftbigg 2 t 3 G parenrightbigg 2. S ( t ) = S 0 sin parenleftbigg 2 π t 3 G parenrightbigg 3. S ( t ) = S 0 sin parenleftbigg 3 t 2 π G parenrightbigg 4. S ( t ) = S 0 sin parenleftbigg t 3 π G parenrightbigg 5. S ( t ) = S 0 sin parenleftbigg 2 t 3 π G parenrightbigg 6. S ( t ) = S 0 sin parenleftbigg t 2 π G parenrightbigg 7. S ( t ) = S 0 sin parenleftbigg 3 π t 2 G parenrightbigg 8. S ( t ) = S 0 sin parenleftbigg 2 π t G parenrightbigg 9. S ( t ) = S 0 sin parenleftbigg 3 t 2 G parenrightbigg 10. S ( t ) = S 0 sin parenleftbigg 3 π t G parenrightbigg correct Explanation: The equation of a wave with zero displace- ment at the origin is given by S ( t ) = S 0 sin( ω t ) = S 0 sin parenleftbigg 2 π T t
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern