Respuesta :
The velocity at mean position is 50 cm/sec
Explanation:
The spring is stretched by a force = 200 x 980 dynes through a length 100 cm . By Hooks law The force F = - k x
here k is spring constant and x is displacement of weight .
Thus 200 x 980 = - k x 100
or k = 1960 dynes/cm
The time period of spring can be found by relation
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
= 2π[tex]\sqrt{\frac{200}{1960} }[/tex] = 2 sec
The frequency of vibration is taken as the reciprocal of time period
Thus frequency ν = [tex]\frac{1}{T}[/tex] = [tex]\frac{1}{2}[/tex] = 0.5 revolution / sec
b. The maximum acceleration is at the end points of vibration , and is equal to acceleration due to gravity .
c. The velocity at mean position can be calculated from the kinetic energy relation of spring .
The kinetic energy of spring = [tex]\frac{1}{2}[/tex] k x²
and it is converted into kinetic energy of mass at mean position
Thus [tex]\frac{1}{2}[/tex] k x² = [tex]\frac{1}{2}[/tex] m v²
or v = [tex]\sqrt{\frac{k}{m} }[/tex] x
= [tex]\sqrt{\frac{1960}{200} }[/tex] x 5 = 50 cm/sec
