Question

The wave function for a standing wave on a string is described by

y(x, t) = 0.016 sin(4πx) cos (57πt),

where y and x are in meters and t is in seconds. Determine the maximum displacement and maximum speed of a point on the string at the following positions.

(a) x = 0.10 m
y_max =  m
v_max =  m/s

(b) x = 0.25 m
y_max =  m
v_max =  m/s

(c) x = 0.30 m
y_max =  m
v_max =  m/s

(d) x = 0.50 m
y_max =  m
v_max =  m/s

2) Two compact sources of sound oscillate in phase with a frequency of 140 Hz where the speed of sound is 340 m/s. At a point that is 3.20 meters from one source and 3.83 meters from the other, the amplitude of the sound from each source separately is A. What is the amplitude of the resultant wave at that point?
✕ A

Answer 1

Now Displacement as a funetion of x. yout 0016 Sm &172) (Sm(5777t) mai 2 LE 0.016x576$m&-wra) (marmitude) 9) @u= a.lom. Ymax = 0.016 XSmd 0-471) =0.0152m max = 0.01645777 XSmCo 450) =2.73mio b@x=0-25m ; Yonex=0-0165m (I) =0 umax's 001645711 x for G120 J@ 2=0.30 m; gmax= 0.0168m (1.27 ) =-0.0094m. Vincex = 0·0165771 Sm (1.27) =-1.68 mio d) @ xzosom i Ymax=0.0/6x Sm (271) =0 Umes 0.016X571TX Snel) 20. 3 Ýapbeed Phase difference is given by the relation = 20 * path ditterence. 08=21AOX ; D = wavelength 17 m. 146 Now path difference. 3 X2-21 3.83-3.20 zo.6 3 n frequeng = 340

04=27X7X0.63 = 20.527 1.7 Now Repultent amplitrede can be calculated by the relation ! AF =AzzA Area = dA2+A2 + 2A6 & =1242C7 Cop (0-5261)) = 1.37A Ano cao -