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MariettaO [177]
3 years ago
12

A child with mass of 25 kg gets into a toy car with mass of 80 kg on a playground, causing it to sink on its springs (with effec

tive spring constant 791 N/m). An adult walks by and gives the top of the car a shove, causing it to undergo oscillations with amplitude 31 cm in the vertical direction. Assuming the oscillations are simple harmonic, what is the frequency of the oscillations?
Physics
1 answer:
Bess [88]3 years ago
4 0

Answer:

The frequency of the oscillation is 0.9Hz

Explanation:

This problem bothers on simple harmonic motion of a spring

Given data

Mass of the child m= 25kg

Spring constant k=791 N/m

Amplitude a= 31cm

But the period of the motion as a result of the adults sholve is expressed as

T=2π√m/k

T=2*3.142√25/791

T=6.284√0.031

T=6.284*0.176

T=1.11 sec

But frequency F=1/T

F=1/1.11

F=0.9Hz

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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
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Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

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