# Solution to HW14 Fall-2002

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2 Since s max µs F N, where F N is the nrmal rce due t the spring pushing perpendicular t the blck (i.e., F N kx) we have that max s µ s F N µ s (kx) Therere, µ s (kx) mg, and we can write µ s mg kx ( ) ( 2.3 kg) 9.80 m/s2 ( 480 N/m) m ( ) 0.92 CJ5 10.P.017. REASONING AND SOLUTION a. Since the bject scillates between ± m, the amplitude the mtin is m. 10.4, b. Frm the graph, the perid is T 4.0 s. Therere, accrding t Equatin ω 2π T 2π 1.6 rad/s 4.0 s ω c. Equatin relates the angular requency t the spring cnstant: k / m. Slving r k we ind k ω 2 m (1.6 rad/s) 2 (0.80 kg) 2.0 N/m d. At t 1.0 s, the graph shws that the spring has its maximum displacement. At this lcatin, the bject is mmentarily at rest, s that its speed is v 0m/s. e. The acceleratin the bject at t 1.0 s is a maximum, and its magnitude is a max Aω 2 (0.080 m)(1.6 rad/s) m/s 2 CJ5 10.P.020. REASONING AND SOLUTION Fr mass 1 we have Fr mass 2 we have Therere, 1 1 2π 2 1 2π 1 2 k (10.11) k + m 2 + m

3 Squaring and slving yields + m r + m Slving r the rati m 2 / gives /m CJ5 10.P.029. SSM REASONING AND SOLUTION I we neglect air resistance, nly the cnservative rces the spring and gravity act n the ball. Therere, the principle cnservatin mechanical energy applies. When the 2.00 kg bject is hung n the end the vertical spring, it stretches the spring by an amunt x, where x F k mg k (2.00 kg)(9.80 m/s2 ) m (10.1) 50.0 N/m This psitin represents the equilibrium psitin the system with the 2.00-kg bject suspended rm the spring. The bject is then pulled dwn anther m and released rm rest (v0 0 m/s). At this pint the spring is stretched by an amunt m m m. This pint represents the zer reerence level (h 0m) r the gravitatinal ptential energy. h 0 m: The kinetic energy, the gravitatinal ptential energy, and the elastic ptential energy at the pint release are: mv 2 0 m 2 KE (0 m/s) 0 J PE mgh mg(0 m) 0 J gravity PE elastic 1 2 kx (50.0 N/m)(0.592 m) J The ttal mechanical energy E 0 at the pint release is the sum the three energies abve: E J. h m: When the bject has risen a distance h m abve the release pint, the spring is stretched by an amunt m m m. Since the ttal mechanical energy is cnserved, its value at this pint is still E 8.76 J. The gravitatinal and elastic ptential energies are: PE gravity mgh (2.00 kg)(9.80 m/s 2 )(0.200 m) 3.92 J

4 PE elastic 1 2 kx2 1 2 (50.0 N/m)(0.392 m) J Since KE + PE gravity + PE elastic E, KE E PE gravity PE elastic 8.76 J 3.92 J 3.84 J 1.00 J h m: When the bject has risen a distance h m abve the release pint, the spring is stretched by an amunt m m m. At this pint, the ttal mechanical energy is still E 8.76 J. The gravitatinal and elastic ptential energies are: PE gravity mgh (2.00 kg)(9.80 m/s 2 )(0.400 m) 7.84 J PE elastic 1 2 kx2 1 2 (50.0 N/m)(0.192 m) J The kinetic energy is KE E PE PE 8.76 J 7.84 J 0.92 J 0 J gravity elastic The results are summarized in the table belw: h KE PE grav PE elastic E m 0.00 J 0.00 J 8.76 J 8.76 J J 3.92 J 3.84 J 8.76 J m 0.00 J 7.84 J 0.92 J 8.76 J Chapter 16 CJ5 16.CQ.004. REASONING AND SOLUTION A wave mves n a string with cnstant velcity. It is nt crrect t cnclude that the particles the string always have zer acceleratin. As Cnceptual Example 3 discusses, it is imprtant t distinguish between the speed the waves n the string, v wave, and the speed the particles in the string, v particle. The wave speed v wave is determined by the prperties the string; namely, the tensin in the string and the linear mass density the string. These prperties determine the speed with which the disturbance travels alng the string. The wave speed will remain cnstant as lng as these prperties remain unchanged.

5 The particles in the string scillate transversely in simple harmnic mtin with the same amplitude and requency as the surce the disturbance. Like all particles in simple harmnic mtin, the acceleratin the particles cntinually changes. It is zer when the particles pass thrugh their equilibrium psitins and is a maximum when the particles are at their maximum displacements rm their equilibrium psitins. CJ5 16.CQ.011. REASONING AND SOLUTION A ludspeaker prduces a sund wave. The sund wave travels rm air int water. As indicated in Table 16.1, the speed sund in water is apprximately ur times greater than it is in air. We are tld in the hint that the requency the sund wave des nt change as the sund enters the water. The relatinship between the requency, the wavelength λ, and the speed v a wave is given by Equatin 16.1: v λ. Since the wave speed increases and the requency remains the same as the sund enters the water, the wavelength the sund must increase. CJ5 16.CQ.017. REASONING AND SOLUTION Accrding t Table 16.1, the speed sund in air at 20 C is v 343 m/s, while its value in water at the same temperature is v 1482 m/s. These values inluence the Dppler eect, because which an bserver hears a requency that is dierent rm the requency s that is emitted by the surce sund. Fr purpses this questin, we assume that s and that the speed at which the surce mves is v s 25 m/s. Our cnclusins, hwever, will be valid r any values s and v s. a. The Dppler-shited requency when the surce appraches the bserver is given by Equatin as s 1 ( v s / v). Applying this equatin r air and water, we ind Air Water 1079 Hz 1 (25 m s )/(343 m s ) 1017 Hz 1 (25 m s )/(1482 m s ) The change in requency due t the Dppler eect in air is greater. b. The Dppler-shited requency when the surce mves away rm the bserver is given by Equatin as s 1 + ( v s / v). Applying this equatin r air and water, we ind Air 932 Hz 1 + (25 m s )/(343 m s ) Water 983 Hz 1 + (25 m s )/(1482 m s )

6 The change in requency due t the Dppler eect in air is greater. CJ5 16.P.014. REASONING AND SOLUTION Initially, the speed the wave is v λ. Ater the adjustment the speed the wave is v 2λ. Dividing, we btain v/v 2. Nw r each case Divisin yields Hence, F (m/l)v 2 and F (m/l)v 2 F/F v 2 /v 2 4 F 4F 4(58 N) 230 N CJ5 16.P.023. SSM REASONING The mathematical rm r the displacement a wave traveling in the x directin is given by Equatin 16.4: y A sin 2π t + 2π x λ. Using Equatin 16.1 and the act that 1/T, we btain the llwing numerical values r and λ : 1/T 1.3 Hz, and λ v / 9.2 m. Omitting units and substituting these values r and λ int Equatin 16.4 gives ( 0.37 m ) sin ( 2.6 π t 0.22 π ) y + x CJ5 16.P.024. REASONING AND SOLUTION We ind rm the upper graph that λ m m m and Α m. Frm the lwer graph we ind that Τ 0.30 s 0.10 s 0.20 s. Then, 1/(0.20 s) 5.0 Hz. Substituting these int Equatin 16.3 we get 2π x y Asin 2 π t and y ( m) sin ( 10 π t 50π x) λ CJ5 16.P.035. REASONING AND SOLUTION a. In rder t determine the rder arrival the three waves, we need t knw the speeds each wave. The speeds r air, water and steel are v a 343 m/s, v w 1482 m/s, v s 5040 m/s The rder arrival is steel wave irst, water wave secnd, air wave third. b. Calculate the length time each wave takes t travel 125 m. Therere, the delay times are t s (125 m)/(5040 m/s) s t w (125 m)/(1482 m/s) s t a (125 m)/(343 m/s) s

7 t 12 t w t s s s s t 13 t a t s s s s CJ5 16.P.052. REASONING AND SOLUTION The intensity the "direct" sund is given by text Equatin 16.9, P I DIRECT 4 πr 2 The ttal intensity at the pint in questin is I TOTAL I DIRECT + I REFLECTED 4 π (3.0 m) W W/m W/m 2 CJ5 16.P.058. REASONING AND SOLUTION We knw β 2 β 1 (10 db) lg (I 2 /I 1 ), s that the rati the intensities can be written as ( ) ( ) ( β β )/ 10dB (61 db 23 db)/ 10 db 2 1 I /I CJ5 16.P.071. REASONING The bserved requency changes because the Dppler eect. As yu drive tward the parked car (a statinary surce sund), the Dppler eect is that given by Equatin As yu drive away rm the parked car, Equatin applies. case: SOLUTION Equatins and give the bserved requency in each ( ) ( + v v, tward s v, away s v Driving tward parked car Driving away rm parked car 1 / and 1 / ) Subtracting the equatin n the right rm the ne n the let gives the change in the bserved requency: 2 v / v, tward, away s Slving r the bserver s speed (which is yur speed), we btain (, tward, away ) ( 343 m/s)( 95 Hz) v v Hz s ( ) 17 m/s

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