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agasfer [191]
3 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]3 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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nlexa [21]

Answer:

ω = 380π rad/s

Explanation:

The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π

ω = 2πf

then

ω = 2π(190)

ω = 380π rad/s

8 0
3 years ago
Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i
Naddik [55]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

7 0
3 years ago
An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,
Taya2010 [7]

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

F=\frac{GM_1M_2}{r^2}

Force F=mg'=\frac{GMm}{4R^2}-----1

Force on earth surface F=mg=\frac{GMm}{R^2}------2

Divide 1 and 2 we get

\frac{g'}{g}=\frac{R^2}{4R^2}

g'=\frac{g}{4}  

7 0
3 years ago
Estimate the average rate of change in elevation from d to points i, ii, and iii, assuming the distance from d to i is 3500 m, t
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Ummm i am not going to be able say i am high

8 0
3 years ago
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g/m^{2} = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = \frac{1}{2L}\sqrt{\frac{T}{M} }

⇒      T = 4L^{2}f^{2}M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × 2^{2} ×65^{2} × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

4 0
3 years ago
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