# sample4 - bhappu(nkb284 Sample Midterm 4 orin(58140 This...

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bhappu (nkb284) – Sample Midterm 4 – forin – (58140) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points A mass attached to a spring executes sim- ple harmonic motion in a horizontal plane with an amplitude o± 2 . 09 m. At a point 1 . 8601 m away ±rom the equilibrium, the mass has speed 2 . 01 m / s. What is the period o± oscillation o± the mass? Consider equations ±or x ( t ) and v ( t ) and use sin 2 + cos 2 = 1 to calculate ω . Correct answer: 2 . 97891 s. Explanation: Let A =2 . 09 m , x =1 . 8601 m , and v . 01 m / s . The simplest solution uses the equation ±or simple harmonic motion x ( t )= A sin( ω t ) , and its time derivative v ( t dx dt = cos( ) . ²or any angle (such as the phase angle ) sin 2 ( ) + cos 2 ( ) = 1 , so ± x A ² 2 + ± v A ω ² 2 , and ± v ω ² 2 = A 2 - x 2 . Consequently, ω = v A 2 - x 2 = 2 . 01 m / s ³ (2 . 09 m) 2 - (1 . 8601 m) 2 . 10922 s - 1 and T = 2 π ω = 2 π 2 . 10922 s - 1 = 2 . 97891 s . 002 10.0 points A block o± unknown mass is attached to a spring o± spring constant 7 . 9N / m and under- goes simple harmonic motion with an ampli- tude o± 9 . 2 cm. When the mass is hal±way between its equilibrium position and the end- point, its speed is measured to be 23 cm / s. Calculate the mass o± the block. Correct answer: 0 . 948 kg. Explanation: Let : k =7 . / m , A =9 . 2 cm , and v = 23 cm / s . I± the maximum displacement (amplitude) is A , the hal±way displacement is A 2 . By energy conservation, K i + U i = F f + U f 0+ 1 2 kA 2 = 1 2 mv 2 + 1 2 k ´ A 2 µ 2 2 = 2 + 1 4 2 m = 3 2 4 v 2 = 3 (7 . / m) (0 . 092 m) 2 4 (0 . 23 m / s) 2 = 0 . 948 kg . 003 10.0 points A hoop ( i.e. , a ring) o± radius 2 . 1 m and mass 14 kg is suspended ±rom a pivot on the perimeter o± the hoop as shown in the Fgure. The acceleration o± gravity is 9 . 8m / s 2 . x y θ

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bhappu (nkb284) – Sample Midterm 4 – forin – (58140) 2 Find the angular ±requency ω o± small os- cillations. Correct answer: 1 . 52753 rad / s. Explanation: Basic Concepts T =2 π ± I mg d Solution: From the parallel axis theorem, the moment o± inertia o± the hoop about the pivot is given by I = I c + Md 2 = mR 2 + 2 2 Hence the angular ±requency is ω = ² I = ² mg R 2 2 = ² g 2 R =1 . 52753 rad / s . 004 10.0 points A0 . 12 kg mass is attached to a spring and undergoes simple harmonic motion with a pe- riod o± 0 . 279 s. The total energy o± the system is 3 . 9 J. Find the amplitude o± the motion. Correct answer: 0 . 357998 m. Explanation: Let : m =0 . 12 kg , T . 279 s , and E =3 . 9J . T π ² m k k = 4 π 2 m T 2 = 4 π 2 (0 . 12 kg) (0 . 279 s) 2 = 60 . 8601 N / m , so the total energy is E = 1 2 kA 2 A = ² 2 E k = ± 2 (3 . 9 J) 60 . 8601 N / m = 0 . 357998 m . 005 10.0 points Earthquakes produce two kinds o± 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 di²erent speeds; the P waves are always ±aster than the S waves, but their exact speeds depend on depth and location. For the pur- pose o± this exercise, we assume the P wave’s speed to be 8790 m / s while the S waves travel at a slower speed o± 4080 m / s.
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sample4 - bhappu(nkb284 Sample Midterm 4 orin(58140 This...

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