A transverse wave on a string is described by the following wavefunction. y = (0

Question

A transverse wave on a string is described by the following wavefunction. y = (0.105 m)sin [(x/10 +5t)] (a) Determine the transverse speed andacceleration at t = 0.260 s forthe point on the string located at x = 1.50 m.
Enter anumber. 1 m/s
Enter anumber. 2 m/s2

(b) What are the wavelength, period, and speed of propagation ofthis wave?
Enter anumber. 3 m
Enter anumber. 4 s
Enter anumber. 5 m/s y = (0.105 m)sin [(x/10 +5t)] (a) Determine the transverse speed andacceleration at t = 0.260 s forthe point on the string located at x = 1.50 m.
Enter anumber. 1 m/s
Enter anumber. 2 m/s2

(b) What are the wavelength, period, and speed of propagation ofthis wave?
Enter anumber. 3 m
Enter anumber. 4 s
Enter anumber. 5 m/s Enter anumber. Enter anumber. Enter anumber. Enter anumber. Enter anumber.

Explanation / Answer

Given that                y = (0.105 m) sin[(x/10 +5t)] a) The transverse speed of the wave is given by differensatingw.r.t t at at t = 0.260 s for thepoint on the string located at x = 1.50 m            v = dy/dt =d((0.105 m) sin[(x/10 +5t)])/dt                           = (0.105 m)(5)cos[(x/10 +5t)]                           = (0.105 m)(5)cos[((1.50)/10 + 5(0.260))]                           =1.64m/s b) Now the acceleration of the wave is            v = dy/dt =d((0.105 m) sin[(x/10 +5t)])/dt                           = (0.105 m)(5)cos[(x/10 +5t)]                           = -(0.105 m)(5)2sin[(x/10 +5t)]                           =  -(0.105m)(5)2sin[((1.50)/10 + 5(0.260))]                           = -2.054m/s2 2) a) Given that atransverse wave on a string is described by the following wavefunction.                 y = (0.105 m) sin[(x/10 +5t)] Comparingthe above with equation                y = Asin(kx+t) Theamplitude of the wave is (A) = 0.105m Frequency() = 5 Thepropagation constant(k) =/10 Now thetime period is given by                     T = 2/                         =2/5                        = 0.4s Speed ofthe wave is                     v =A                        = (0.105)(5)                        =1.6485m/s Wavelength of the wave is                     k =2/ Then = 2/k             =2/(/10)            = 20m b) Now the acceleration of the wave is            v = dy/dt =d((0.105 m) sin[(x/10 +5t)])/dt                           = (0.105 m)(5)cos[(x/10 +5t)]                           = -(0.105 m)(5)2sin[(x/10 +5t)]                           =  -(0.105m)(5)2sin[((1.50)/10 + 5(0.260))]                           = -2.054m/s2 2) a) Given that atransverse wave on a string is described by the following wavefunction.                 y = (0.105 m) sin[(x/10 +5t)] Comparingthe above with equation                y = Asin(kx+t) Theamplitude of the wave is (A) = 0.105m Frequency() = 5 Thepropagation constant(k) =/10 Now thetime period is given by                     T = 2/                         =2/5                        = 0.4s Speed ofthe wave is                     v =A                        = (0.105)(5)                        =1.6485m/s Wavelength of the wave is                     k =2/ Then = 2/k             =2/(/10)            = 20m 2) a) Given that atransverse wave on a string is described by the following wavefunction.                 y = (0.105 m) sin[(x/10 +5t)] Comparingthe above with equation                y = Asin(kx+t) Theamplitude of the wave is (A) = 0.105m Frequency() = 5 Thepropagation constant(k) =/10 Now thetime period is given by                     T = 2/                         =2/5                        = 0.4s Speed ofthe wave is                     v =A                        = (0.105)(5)                        =1.6485m/s Wavelength of the wave is                     k =2/ Then = 2/k             =2/(/10)            = 20m                 y = (0.105 m) sin[(x/10 +5t)] Comparingthe above with equation                y = Asin(kx+t) Theamplitude of the wave is (A) = 0.105m Frequency() = 5 Thepropagation constant(k) =/10 Now thetime period is given by                     T = 2/                         =2/5                        = 0.4s Speed ofthe wave is                     v =A                        = (0.105)(5)                        =1.6485m/s Wavelength of the wave is                     k =2/ Then = 2/k             =2/(/10)            = 20m

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