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agasfer [191]
2 years ago
10

Define Simple harmonic motion (SHM). A body moving with SHM has an amplitude of 10 cm and a frequency of 100 Hz. Find (a) the pe

riod of oscillation (b) the acceleration at the maximum displacement (c) the velocity at the centre of motion.
Physics
1 answer:
luda_lava [24]2 years ago
7 0

Answer:

Explanation:

Simple harmonic motion is the motion that involves the cyclic and periodic motion of an object about a central point or equilibrium position, due to the effect of a restoring force, such that the object has the same maximum displacement during its motion on either side of the equilibrium point

(a) The parameters of the body moving with SHM are;

The amplitude of the motion of the body, a = 10 cm

The frequency of the complete cycles of the motion of the body, f = 100 Hz

The period of the oscillation, 'T', is the time it takes o complete one cycle

T = 1/f

∴ T = 1/100 Hz = 0.01 seconds

(b) The acceleration, a = -A·ω²·sin(ω·t)

Where;

One complete oscillation is given as ω·t = 2·π

Therefore at either end at the maximum displacement, which is at either end of the equilibrium position is reached at half a complete cycle, or ω·t = 2·π/2 = π

At the maximum displacement a = -A·ω²·sin(ω·t) = A·ω²·sin(π) = 0

Therefore, we have;

The acceleration, of the body at the maximum displacement, a = 0 m/s².

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If a red blood cell is 7 x 10^-^6 m in size, it means that such a cell is 7 microns or micrometer in size. The unit that completes the sentence will, thus, be microns or micrometer.

According to international system of measurement:

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  • 10^{-2} m = centimeter
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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
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We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

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Answer:

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Answer:

A

Explanation:

I TOOK IT

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