Question

Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from largest to smallest.
m = 2 kg , k = 2 N/m

m = 2 kg , k = 4 N/m

m = 4 kg , k = 2 N/m

m = 1 kg , k = 4 N/m

Answer 1

The periods of oscillation for the mass–spring systems from largest to smallest is:

1. m = 4 kg , k = 2 N/m (T = 8.89 s)
2. m = 2 kg , k = 2 N/m (T = 6.28 s)
3. m = 2 kg , k = 4 N/m (T = 4.44 s)
4. m = 1 kg , k = 4 N/m (T = 3.14 s)

Explanation:

The period of oscillation in a simple harmonic motion is defined as the following formulation:

$T=2\pi \sqrt{\frac{m}{k} }$

Where:

T = period of oscillation

m = inertia mass of the oscillating body

k = spring constant

m = 2 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{2}{2} }$

$T=2\pi$

T = 6.28 s

m = 2 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{2}{4} }$

$T=2\pi \sqrt{\frac{1}{2} }$

T = 4.44 s

m = 4 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{4}{2} }$

$T=2\pi \sqrt{2 }$

T = 8.89 s

m = 1 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{1}{4} }$

$T=\pi$

T = 3.14 s

Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:

1. m = 4 kg , k = 2 N/m (T = 8.89 s)
2. m = 2 kg , k = 2 N/m (T = 6.28 s)
3. m = 2 kg , k = 4 N/m (T = 4.44 s)
4. m = 1 kg , k = 4 N/m (T = 3.14 s)

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Answer 2

Answer:

System 3:

m = 4 kg , k = 2 N/m

T=8.89s

System 1

m = 2 kg , k = 2 N/m

T=6.28s

System 2:

m = 2 kg , k = 4 N/m

T=4.44s

System 4:

m = 1 kg , k = 4 N/m

T=3.14s

Explanation:

The period of oscillation in a simple harmonic motion is defined by:

$T=2\pi \sqrt{\frac{m}{k} }$

Where:

$T=Period\hspace{3}of\hspace{3}oscillation\\m=Inertial\hspace{3}mass\hspace{3}of\hspace{3}the\hspace{3}oscillating\hspace{3}body\\k= Spring\hspace{3}constant$

Now, let:

$T_1=Period\hspace{3}of\hspace{3}oscillation\hspace{3}associated\hspace{3}to\hspace{3}system\hspace{3}1\\T_2=Period\hspace{3}of\hspace{3}oscillation\hspace{3}associated\hspace{3}to\hspace{3}system\hspace{3}2\\T_3=Period\hspace{3}of\hspace{3}oscillation\hspace{3}associated\hspace{3}to\hspace{3}system\hspace{3}3\\T_4=Period\hspace{3}of\hspace{3}oscillation\hspace{3}associated\hspace{3}to\hspace{3}system\hspace{3}4$

For $T_1$ :

$T_1=2\pi \sqrt{\frac{2}{2} } =2\pi\approx6.28s$

For $T_2$ :

$T_2=2\pi \sqrt{\frac{2}{4} } \approx4.44s$

For $T_3$ :

$T_3=2\pi \sqrt{\frac{4}{2} } \approx8.89s$

For $T_4$ :

$T_4=2\pi \sqrt{\frac{1}{4} } =\pi\approx3.14s$

Therefore:

$T_3>T_1>T_2>T_4$