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Dmitry_Shevchenko [17]
3 years ago
14

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x, t) = (5

.60 cm) sin [(0.0340 rad/cm) x] sin [(50.0 rad/s) t], where the origin is at the left end of the string, the x-axis is along the string and the y-axis is perpendicular to the string.
a) Draw a sketch that shows the standing wave pattern.
(b) Find the amplitude of the two travelling waved that make up this standing wave.
(c) What is the length of the string?
(d) Find the wavelength, frequency, period and speed of the travelling waves.
(e) Find the maximum transverse speed of a point on the string.
(f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
Physics
1 answer:
barxatty [35]3 years ago
4 0

Answer:

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?

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Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

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From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

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              t__{A}} = \frac{l}{v__{A}}}

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=>           t__{A}} =  50 \  s

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             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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