The periods of oscillation for the mass–spring systems from largest to smallest is:
The period of oscillation in a simple harmonic motion is defined as the following formulation:
[tex]T=2\pi \sqrt{\frac{m}{k} }[/tex]
Where:
T = period of oscillation
m = inertia mass of the oscillating body
k = spring constant
m = 2 kg , k = 2 N/m
[tex]T=2\pi \sqrt{\frac{2}{2} }[/tex]
[tex]T=2\pi[/tex]
T = 6.28 s
m = 2 kg , k = 4 N/m
[tex]T=2\pi \sqrt{\frac{2}{4} }[/tex]
[tex]T=2\pi \sqrt{\frac{1}{2} }[/tex]
T = 4.44 s
m = 4 kg , k = 2 N/m
[tex]T=2\pi \sqrt{\frac{4}{2} }[/tex]
[tex]T=2\pi \sqrt{2 } [/tex]
T = 8.89 s
m = 1 kg , k = 4 N/m
[tex]T=2\pi \sqrt{\frac{1}{4} }[/tex]
[tex]T=\pi [/tex]
T = 3.14 s
Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:
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The time period for case 3 is largest followed by case 1, case 2 and case 4 . So, the ranking is of the order, [tex]T_{3} > T_{1} >T_{2}> T_{4}[/tex].
The general formula for the time period of mass-spring system is given as,
[tex]T= 2\pi\sqrt{\dfrac{m}{k} }[/tex]
Here, m is the mass of spring and k is the spring constant.
Case 1 - For m = 2kg and k =2 N/m, the time period is,
[tex]T_{1}= 2\pi\sqrt{\dfrac{m}{k} }\\T_{1}= 2\pi\sqrt{\dfrac{1}{1} }\\T_{1}= 6.28 \;\rm s[/tex]
Case 2 - For m = 2kg and k =4 N/m, the time period is,
[tex]T_{2}= 2\pi\sqrt{\dfrac{m}{k} }\\T_{2}= 2\pi\sqrt{\dfrac{2}{4} }\\T_{2}= 4.44 \;\rm s[/tex]
Case 3 - For m = 4 kg and k =2 N/m, the time period is,
[tex]T_{3}= 2\pi\sqrt{\dfrac{m}{k} }\\T_{3}= 2\pi\sqrt{\dfrac{4}{2} }\\T_{3}= 8.88 \;\rm s[/tex]
Case 4 - For m = 1 kg and k =4 N/m, the time period is,
[tex]T_{4}= 2\pi\sqrt{\dfrac{m}{k} }\\T_{4}= 2\pi\sqrt{\dfrac{1}{4} }\\T_{4}= 3.14 \;\rm s[/tex]
Thus, we can conclude that time period for case 3 is largest and for case 4 is smallest. And the ranking is of the order,
[tex]T_{3} > T_{1} >T_{2}> T_{4}[/tex].
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