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Free_Kalibri [48]
3 years ago
14

An object in simple harmonic motion has an amplitude of 0.360 m and an oscillation period of 0.620 s. Determine the maximum spee

d of the motion.
Physics
1 answer:
steposvetlana [31]3 years ago
5 0

Answer:

3.648 m/s

Explanation:

Data provided in the question:

Amplitude, A = 0.360 m

oscillation period, T = 0.620 s

The maximum speed of the motion, V = A × ω

where,

ω = Angular speed

also,

ω = (2π) / T

on substituting the respective values, we get

ω = (2π) / 0.620

or

ω = 10.134 rad / s

therefore,

V = 0.360 × 10.134 = 3.648 m/s

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Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

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What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

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