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2 years ago

The acceleration of positive performing SHM is 12cm/sec at distance of 3cm from the mean position its time period is?

Physics

1 answer:

mina [271]2 years ago

4 0

Answer:

0.25 seconds

Explanation:

this is because speed is measured in metres per second

speed = distance/ time

12cm/sec = 3cm/T

T = 3cm ÷ 12cm/sec

T = 0.25sec

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Answer:

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Answer:

a) x=0 %T=0, b) x= A %T=100%, c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

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x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

% T = 0

b) for x = A

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cos tea = x / A = 1

therefore the angles tea = 2π rad

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% T = 100%

c) the point of maximum compression x = -A

let's look for the angles

cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

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Answer:

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Explanation:

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The acceleration of positive performing SHM is 12cm/sec at distance of 3cm from the mean position its time period is?​

The acceleration of positive performing SHM is 12cm/sec at distance of 3cm from the mean position its time period is?