A 0.156-kg ball is hung vertically from a spring. The spring stretches by 4.18 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position

[tex]= 2 \pi \sqrt{\frac{0.156}{3.65}}\\\\= 2 \times 3.14 \sqrt{\frac{0.156}{3.65}}\\\\= 6.28 \sqrt{\frac{0.156}{3.65}}\\\\=6.28 \times 0.108210508\\\\=0.67956199\\\\=0.68\\\\[/tex]

The time to calculating the 4 oscillation:

[tex]\to 4T=4\times 0.68 =2.72 \ seconds \\\\[/tex]

shubhamchouhanvrVT shubhamchouhanvrVT · Answer 2 · 2022-04-06T11:05:19+02:00

The time taken by the spring to complete 4 complete oscillations will be t=2.72 sec

What are oscillations?

Oscillation is the process of repeating variation of the quantity or the recitative or periodic variations with respect to the time.

Here it is given that

mass =0.156 kg

stretch x= 4.18 cm

Now from the formula

[tex]kx=mg[/tex]

[tex]k= \dfrac{0.156\times 9.81}{0.418}[/tex]

[tex]k=3.65\ \frac{N}{m}[/tex]

Now the time period will be calculated as:

[tex]T=2\pi\sqrt{\dfrac{m}{k}[/tex]

[tex]T=2\pi\sqrt{\dfrac{0.156}{3.65}[/tex]

[tex]T=6.28\times 0.108210508[/tex]

[tex]T=0.68\ sec[/tex]

Now for the 4 oscillations the total time will be

[tex]T=4\times 0.68=2.72\ sec[/tex]

Hence the time taken by the spring to complete 4 complete oscillations will be t=2.72 sec

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