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Delicious77
17 days 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:
kicyunya [2.2K]17 days ago
7 0

Answer:

a) factor b=\sqrt{2}

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

c) factor b=1

d) factor b=1

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

T=\frac{1}{f}

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

where:

f= represents the frequency of oscillation

m= signifies the mass linked to the spring

k= is the spring's stiffness constant

a) If the mass is doubled:

  • New mass, m'=2m

Thus, the new time period:

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

this leads to factor b=\sqrt{2} as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant, k'=4k

Thus, the new time period:

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

this results in factor b=\frac{1}{2} as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness, k'=4k

New mass, m'=4m

Thus, the new time period:

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

which leads to factor b=1 as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor b=1

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1 month ago
A mechanical high-speed bat is flying along a path perpendicular to a wall. it emits a sound with a frequency f0 . the night is
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28 days ago
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Softa [2035]

Answer:

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Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

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n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

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ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

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When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
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Answer:

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P_2 = \frac{nrT_2}{v}

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let the final pressure be P and the temperature T_1 {f} and T_2 {f}

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A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

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