A string of length 1 m fixed one end and on the other end a block of mass `M = 4 kg` is suspended. The string is set into vibration and represented by equation `y =6 sin ((pi x)/(10)) cos (100 pi t)` where x and y are in cm and t is in seconds.
(i) Find the number of loops formed in the string.
(ii) Find the maximum displacement of a point at `x = 5//3 cm`
(iii) Calculate the maximum kinetic energy of the string.
(iv) Write down the equations of the component waves whose superposition given the wave.
.

Answer

(i) 10
(ii) 3 cm
(iii) 36 J
(iv) `y_(1)=3sin((pix)/(10)-100pit),y_(2)=3sin((pix)/(10)+100pit)`

Explanation

`(2pi)/(lamda)=(pix)/(10)rArr lamda=20cm`
(i) Total no of wavelength `=(L)/(lamda)=(100)/(20)=5`
`:.` Number of loops formed `=2xx 5=10`
(ii) Maximum displacement at `x = (5)/(3)`
`A=6sin((pi)/(10)xx(5)/(3))=3cm`
(iii) `y=6sin((pix)/(10))cos(100pit)`
`=6sin(10pix)cos(100pit)`
`v=6pisin(10pix)sin (100 pit)m//s`
`:.KE_(max)=underset(0)overset(1)(int)(1)/(2)muv^(2)dx=(1)/(2)mu underset(0)overset(4)(int)6pisin(10pix)]^(2) dx`
where `sqrt((T)/(mu))=(100pi)/(10pi)=10=mu=0.4:. KE_(max)=36J`
(iv) `y=6 sin((pix)/(10))cos(100pit)=y_(1)+y_(2)`
`rArr y_(1)=3sin((pix)/(10)-100t),y_(2)=3sin((pix)/(10)+100 pit)`.