Answer:
the spring coefficient is
k=16N/m
Explanation:
Hooks law states that provided the elasticity of a material is not exceeded the extension e is proportional to the applied force
Step one
Analysis of the problem
From analysis of the problem
The mass has a potential energy due to the height it was dropped from, the potential energy is then stored in the spring since it was dropped on the spring which compresses it by 0.5m
Step two
Data
Mass of object m=0.2kg
Height of building =10m
Compression of spring e=0.5m
Spring constant k=?
Step three
According to the principle of energy conservation
mgh=1/2(k*e^2)
Making k subject of formula we have
k=2mgh/e^2
Substituting our data into the expression to get k
Assuming g=9.81m/s
k=2*0.2*10/0.5^2
k=4/0.25
k=16N/m
I'm guessing D. toothpick, but i might be wrong.
Answer:
The value of g is ![g =76.2 m/s^2](https://tex.z-dn.net/?f=g%20%3D76.2%20m%2Fs%5E2)
Explanation:
From the question we are told that
The mass of the weight is ![m = 1.30 kg](https://tex.z-dn.net/?f=m%20%3D%20%201.30%20kg)
The spring constant ![k = 1.73 g/m = 1.73 *10^{-3} \ kg/m](https://tex.z-dn.net/?f=k%20%3D%20%201.73%20g%2Fm%20%3D%201.73%20%2A10%5E%7B-3%7D%20%5C%20kg%2Fm)
The second harmonic frequency is ![f = 100 \ Hz](https://tex.z-dn.net/?f=f%20%3D%20%20100%20%5C%20Hz)
The number of oscillation is ![N = 200](https://tex.z-dn.net/?f=N%20%20%3D%20%20200)
The time taken is ![t = 315 \ s](https://tex.z-dn.net/?f=t%20%3D%20%20315%20%5C%20s)
Generally the frequency is mathematically represented as
![f = \frac{v}{\lambda}](https://tex.z-dn.net/?f=f%20%3D%20%20%5Cfrac%7Bv%7D%7B%5Clambda%7D)
At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated
![l = \lambda](https://tex.z-dn.net/?f=l%20%20%3D%20%20%5Clambda)
Noe from the question the vibrating string is just half of the length of the main string so
Let assume the length of the main string is ![L](https://tex.z-dn.net/?f=L)
So ![l = \frac{L}{2}](https://tex.z-dn.net/?f=l%20%3D%20%20%5Cfrac%7BL%7D%7B2%7D)
The velocity of the vibrating string is mathematically represented as
![v = \sqrt{\frac{T}{\mu} }](https://tex.z-dn.net/?f=v%20%20%3D%20%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D)
Where T is the tension on the string which can be mathematically represented as
![T = mg](https://tex.z-dn.net/?f=T%20%20%3D%20mg)
So
![v = \sqrt{\frac{mg}{k} }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Csqrt%7B%5Cfrac%7Bmg%7D%7Bk%7D%20%7D)
Then
![f = \frac{v}{\frac{L}{2} }](https://tex.z-dn.net/?f=f%20%3D%20%20%5Cfrac%7Bv%7D%7B%5Cfrac%7BL%7D%7B2%7D%20%7D)
=> ![v = \frac{fL }{2}](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7BfL%20%7D%7B2%7D)
=> ![\sqrt{\frac{mg}{k} } = \frac{fL}{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7Bmg%7D%7Bk%7D%20%7D%20%3D%20%20%5Cfrac%7BfL%7D%7B2%7D)
=> ![g = \frac{f^2 L^2 \mu}{4m}](https://tex.z-dn.net/?f=g%20%20%3D%20%20%5Cfrac%7Bf%5E2%20L%5E2%20%5Cmu%7D%7B4m%7D)
substituting values
![g = \frac{(100) * (1.73 *10^{-3} )}{(4 * 1.30)} L^2](https://tex.z-dn.net/?f=g%20%3D%20%20%5Cfrac%7B%28100%29%20%2A%20%281.73%20%2A10%5E%7B-3%7D%20%29%7D%7B%284%20%2A%201.30%29%7D%20%20L%5E2)
![g = 3.326 m^{-1} s^{-2} L^2](https://tex.z-dn.net/?f=g%20%3D%20%20%203.326%20%20m%5E%7B-1%7D%20s%5E%7B-2%7D%20L%5E2)
Generally the period of oscillation is mathematically represented as
![T_p = 2 \pi \sqrt{\frac{L}{g} }](https://tex.z-dn.net/?f=T_p%20%20%3D%20%202%20%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%20%7D)
=> ![L = \frac{T^2 g}{4 \pi ^2}](https://tex.z-dn.net/?f=L%20%20%3D%20%20%5Cfrac%7BT%5E2%20g%7D%7B4%20%5Cpi%20%5E2%7D)
The period can be mathematically evaluated as
![T_p = \frac{t}{N}](https://tex.z-dn.net/?f=T_p%20%20%3D%20%20%5Cfrac%7Bt%7D%7BN%7D)
substituting values
![T_p = \frac{315}{200}](https://tex.z-dn.net/?f=T_p%20%20%3D%20%20%5Cfrac%7B315%7D%7B200%7D)
![T_p = 1.575 \ s](https://tex.z-dn.net/?f=T_p%20%20%3D%201.575%20%5C%20s)
Therefore
![L = \frac{1.575^2 * g }{4 \pi ^2}](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7B1.575%5E2%20%2A%20g%20%7D%7B4%20%5Cpi%20%5E2%7D)
![L = 0.0628 ^2 g](https://tex.z-dn.net/?f=L%20%3D%200.0628%20%5E2%20g)
so
![g = 3.326 m^{-1} s^{-2} L^2](https://tex.z-dn.net/?f=g%20%3D%20%20%203.326%20%20m%5E%7B-1%7D%20s%5E%7B-2%7D%20L%5E2)
substituting for L
![g = 3.326 ((0.0628) g)^2](https://tex.z-dn.net/?f=g%20%3D%20%20%203.326%20%20%20%28%280.0628%29%20g%29%5E2)
=> ![g = \frac{1}{(3.326)* (0.0628)^2}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7B1%7D%7B%283.326%29%2A%20%280.0628%29%5E2%7D)
![g =76.2 m/s^2](https://tex.z-dn.net/?f=g%20%3D76.2%20m%2Fs%5E2)