Answer:
191.36 N/m
Explanation:
From the question,
The Potential Energy of the safe = Energy of the spring when it was compressed.
mgh = 1/2ke²............... Equation 1
Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression
Making k the subject of the equation,
k =2mgh/e²................ Equation 2
Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m
Constant: g = 9.8 m/s²
Substitute into equation 2
k = 2(1100)(9.8)(0.0024)/0.52²
k = 51.744/0.2704
k = 191.36 N/m
Hence the spring constant of the heavy-duty spring = 191.36 N/m
Answer:
Explanation:
As we know that amplitude of forced oscillation is given as
here we know that natural frequency of the oscillation is given as
here mass of the object is given as
angular frequency of applied force is given as
now we have
Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.
