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Paraphin [41]
1 year ago
12

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

d change? That is, if the original period was T and the new period is b⁢T, what is b? It is useful to write out the expression for the period and ask yourself what would happen if you doubled the mass.b. If, instead, you quadruple the spring stiffness but keep the mass the same, what is the factor b?c. If, instead, you quadruple the mass and also quadruple the spring stiffness, what is the factor b?d. If, instead, you quadruple the amplitude (keeping the original mass and spring stiffness), what is the factor b?
Physics
1 answer:
Katena32 [7]1 year ago
7 0

Answer:

a) factor b=\sqrt{2}

b) factor b=\frac{1}{2}

c) factor b=1

d) factor b=1

Explanation:

Time period of oscillating spring-mass system is given as:

T=\frac{1}{f}

T={2\pi} \sqrt{\frac{m}{k} }

where:

f= frequency of oscillation

m= mass of the object attached to the spring

k= stiffness constant of the spring

a) <u>On doubling the mass:</u>

  • New mass, m'=2m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k} }

T'=2\pi\sqrt{\frac{2m}{k} }

T'=\sqrt{2}\times  2\pi\sqrt{\frac{m}{k} } }

T'=\sqrt{2} \times T

where the factor b=\sqrt{2} as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, k'=4k

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T

where the factor  b=\frac{1}{2}  as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, k'=4k

New mas, m'=4m

<u>Then the new time period:</u>

T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times  2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T

where factor b=1 as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor b=1

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Answer:

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n=0.45

As we know that

Lateral strain

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\varepsilon _t=\dfrac{32-30}{30}

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We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

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<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
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In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

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Answer:

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Explanation:

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- The total time taken to cover distance(d) by ben = t_b

Find:

How long will it be before Amanda catches up to Ben?

Solution:

- The distance d traveled by Ben:

                                 d = v_b*t_b

                                 d = 3*t_b

- The distance d traveled by Amanda:

                                 d = v_a*t_a

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                                  t_a = t_b - 1.5

- Substitute the time relationship in distance relationship:

                                  3*t_b = 6*(t_b - 1.5)

                                  3*t_b = 6*1.5

                                      t_b = 2*1.5 = 3 h

- Hence, It would take 3 hours since Ben starts walking that amanda catches up.

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2 years ago
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