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 oladayoademosu · Answer 1 · 2022-03-03T12:02:27+01:00

It takes the system 1.58 s to complete 2.5 oscillations

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

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 = 0.2π s

T = 0.2 × 3.142

T = 0.6284 s

T ≅ 0.63 s

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

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