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A 7. 5 kg block is attached to a horizontally-mounted spring with a spring constant of 750 j/m2. The block is released from rest with the spring extended by 15 cm from its normal length of 27 cm. The horizontal surface along which the block moves is frictionless. How long (in s) will it take this system to complete 2. 5 oscillations?.

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oladayoademosu

It takes the system 1.58 s to complete 2.5 oscillations

Period of oscillation

The period of oscillation of the horizontally mounted spring is T = 2π√(m/k) where

  • m = mass of block = 7.5 kg and
  • k = spring constant = 750 N/m²

Substituting the values of the variables into the equation, we have

T = 2π√(m/k)

T = 2π√(7.5 kg/750 N/m²)

T = 2π√(0.01 kgN/m²)

T = 2π × 0.1 s

T = 0.2Ï€ s

T = 0.2 × 3.142

T = 0.6284 s

T ≅ 0.63 s

Time to complete 2.5 oscillations

Since one oscillation = period = 0.63 s, then 2.5 oscillations = 2.5 × 1 oscillation = 2.5 × 0.63 s

= 1.575 s

≅ 1.58 s

So, it takes the system 1.58 s to complete 2.5 oscillations

Learn more about time to complete oscillations here:

https://brainly.com/question/14961716

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