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zhenek [66]
3 years ago
11

A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl

ock has a displacement of -0.70m, a velocity of -0.80 m/s, and an acceleration of +2.7 m/s2. What is the amplitude of the motion?
Physics
1 answer:
Genrish500 [490]3 years ago
3 0

Answer:

Explanation:

Given

mass of block m=5.7\ kg

at t=0 s

displacement is x=-0.7\ m

velocity v=-0.8\ m/s

acceleration a=2.7\ m/s^2

suppose x=A\cos (\omega t+\phi )   is the general equation of SHM

where A=amplitude

\omega=natural frequency of oscillation

therefore velocity and acceleration is given by

v=-A\omega \sin (\omega t+\phi )

a=A\omega ^2\cos (\omega t+\phi )

for t=0

-0.7=A\cos (\phi )---1

v=-0.8=-A\omega \sin(\phi)---2

a=2.7=-A\omega ^2\cos(\phi )----3

divide 1 and 3 we get

\omega ^2=\frac{27}{7}

\omega =\sqrt{\frac{27}{7}}

Now square and 1 and 2 we get

(0.7)^2+(\frac{0.8}{\omega })^2=A^2

A^2=0.49+0.166

A=0.81\ m

     

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