The periods of oscillation for the mass–spring systems from largest to smallest is:
- m = 4 kg , k = 2 N/m (T = 8.89 s)
- m = 2 kg , k = 2 N/m (T = 6.28 s)
- m = 2 kg , k = 4 N/m (T = 4.44 s)
- m = 1 kg , k = 4 N/m (T = 3.14 s)
<h3>Explanation:</h3>
The period of oscillation in a simple harmonic motion is defined as the following formulation:
![T=2\pi \sqrt{\frac{m}{k} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D)
Where:
T = period of oscillation
m = inertia mass of the oscillating body
k = spring constant
m = 2 kg , k = 2 N/m
![T=2\pi \sqrt{\frac{2}{2} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B2%7D%7B2%7D%20%7D)
![T=2\pi](https://tex.z-dn.net/?f=T%3D2%5Cpi)
T = 6.28 s
m = 2 kg , k = 4 N/m
![T=2\pi \sqrt{\frac{2}{4} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B2%7D%7B4%7D%20%7D)
![T=2\pi \sqrt{\frac{1}{2} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%20%7D)
T = 4.44 s
m = 4 kg , k = 2 N/m
![T=2\pi \sqrt{\frac{4}{2} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B4%7D%7B2%7D%20%7D)
![T=2\pi \sqrt{2 }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B2%20%7D%20)
T = 8.89 s
m = 1 kg , k = 4 N/m
![T=2\pi \sqrt{\frac{1}{4} }](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B1%7D%7B4%7D%20%7D)
![T=\pi](https://tex.z-dn.net/?f=T%3D%5Cpi%20)
T = 3.14 s
Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:
- m = 4 kg , k = 2 N/m (T = 8.89 s)
- m = 2 kg , k = 2 N/m (T = 6.28 s)
- m = 2 kg , k = 4 N/m (T = 4.44 s)
- m = 1 kg , k = 4 N/m (T = 3.14 s)
Learn more about simple harmonic motion brainly.com/question/13058166
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You are dealing with pulleys?
can be done with addition of the two equations to eliminate T.
![(2^{*}) \quad {} -m_1 g \cos \theta + T = m_1 a _2](https://tex.z-dn.net/?f=%282%5E%7B%2A%7D%29%20%5Cquad%20%7B%7D%20-m_1%20g%20%5Ccos%20%5Ctheta%20%2B%20T%20%3D%20m_1%20a%20_2)
+
![(4^{*}) \quad 3m_1 g - T = 3m_1 a_2](https://tex.z-dn.net/?f=%284%5E%7B%2A%7D%29%20%5Cquad%203m_1%20g%20-%20T%20%3D%203m_1%20a_2%20)
=
![(-m_1 g \cos\theta + 3m_1 g) + (T - T) = m_1 a _2 + 3m_1 a_2 \implies \\ \\ (-m_1 g \cos\theta + 3m_1 g) = 4m_1 a_2 \implies \\ \\ m_1(- g \cos\theta + 3g)= 4m_1 a_2](https://tex.z-dn.net/?f=%28-m_1%20g%20%5Ccos%5Ctheta%20%20%2B%20%203m_1%20g%29%20%2B%20%28T%20-%20T%29%20%3D%20m_1%20a%20_2%20%2B%203m_1%20a_2%20%5Cimplies%20%5C%5C%20%5C%5C%0A%28-m_1%20g%20%5Ccos%5Ctheta%20%20%2B%20%203m_1%20g%29%20%3D%204m_1%20a_2%20%5Cimplies%20%5C%5C%20%5C%5C%0Am_1%28-%20g%20%5Ccos%5Ctheta%20%20%2B%20%203g%29%3D%204m_1%20a_2%20)
we can cancel m₁ by dividing both sides by it, assuming mass is not zero
![(- g \cos\theta + 3g)= 4a_2 \implies \\ \\ a_2 = \dfrac{- g \cos\theta + 3g}{4} \\ \\ a_2 = \dfrac{- 9.80 \cos 60 + 3(9.80)}{4} \\ \\ a_2 = 6.125 \text{m/s}^2 ](https://tex.z-dn.net/?f=%28-%20g%20%5Ccos%5Ctheta%20%2B%203g%29%3D%204a_2%20%5Cimplies%20%5C%5C%20%5C%5C%0Aa_2%20%3D%20%5Cdfrac%7B-%20g%20%5Ccos%5Ctheta%20%2B%203g%7D%7B4%7D%20%5C%5C%20%5C%5C%0Aa_2%20%3D%20%5Cdfrac%7B-%209.80%20%20%5Ccos%2060%20%2B%203%289.80%29%7D%7B4%7D%20%5C%5C%20%5C%5C%0Aa_2%20%3D%206.125%20%5Ctext%7Bm%2Fs%7D%5E2%0A)
a₂ = 6.125 m/s² ( do significant digits if you need to)
Answer:
Explanation:
Given
Force=18lb
extension=8in
Using Hooke's law to get the spring constant(k)
F=ke
Then,
K=f/e
K=18/8
K=2.25lb/in
Work done by spring is given by
W=1/2Fe
Or W=1/2ke²
Then,
Work done in stretching the spring to 14in
W=1/2ke²
W=0.5×2.25×14²
W=220.5lbin
1 Inch-pounds Force to Joules = 0.113J
Then, to joules
W=0.133×220.5
W=29.33J
Answer:
Going with the current. You every watch Finding Nemo and the turtle guy says " Going with the waves bro"
Explanation: