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A block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The other end of the spring is attached to a wall. The block is pulled away from the spring’s unstrained position by a distance x0 = 0.050 m and released from rest. The period of the subsequent periodic motion of the block is 0.64 s. At what distance from the unstrained position is the speed of the block equal to 0.30 m/s?

Respuesta :

Answer:

X = .04 m

Explanation:

The angular velocity ω = 2π / T

= 2 X 3.14 / .64 = 9.8 radian per second.

Amplitude of oscillation A = .05 m

Maximum velocity V₀ = ω A

= 9.8 X .05 = .49 m/s

If V be velocity at distance X then

V = ω √( A² - X² )

V = .3 m/s at X

Putting the values

[tex].3 =9.8\sqrt{.05^2 - X^2}[/tex]

[tex]\sqrt{.05^2-X^2} =    \frac{.3}{9.8} }[/tex]

X = .04 m

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