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Paraphin [41]
2 years 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]2 years 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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A bowling ball of mass m=1.7kg is launched from a spring compressed by a distance d=0.31m at an angle of theta=37 measured from
vodomira [7]

Answer:

k = 1 700.7 N/m

v0 = 9.8 m/s^2

Explanation:

Hello!

We can answer this question using conservation of energy.

The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.

When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.

PS = (1/2) k x^2  <em>where x is the compresion or elongation of the spring</em>

PG = mgh

a)

Since energy must be conserved and we are neglecting any energy loss:

PS = PG

Solving for k

k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)

k = 1 700.7 N/m

b)

Since the potential energy of the spring transfors to kinetic energy of the ball we have that:

PS = KE

that is:

(1/2) k x^2 = (1/2) m v0^2

Solving for v0

v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)

v0 = 9.8 m/s^2

8 0
2 years ago
You decide it is time to clean your pool since summer is quickly approaching. Your pool maintenance guide specifies that the chl
kotykmax [81]

Answer:

Concentration of Cl2: 1,048 ppm

Explanation:

The unit of measurement molarity (M) represents the same magnitud as mole/L (number of moles of the substance in one liter of solution). The concentration in ppm represents the same as mg/L (number of miligrammes of the substance in one liter of solution). In order to convert from M to ppm we have to consider, then, mass and volume. We can see that in both cases the volume is expressed in liters, so we don't have to do anything to change it. The problem comes when you see that you have to convert moles, from the molatiry, to miligrammes, to get the result in mg/L, meaning ppm.

First of all, notice that the substance is Cl2, so we need to find a relationship between the number of moles and its mass. If you look in the periodic table you'll see that the atomic mass for Cl is 35,4 g/mole (grammes of Cl in one mole). So, there is a way to relate the moles to the mass of the substance and it is represented on the equation below:

mass=mole*atomic mass

We want to find the mass (m) and we know the amount of moles of Cl2 in the solution (moles=2,96x10^-5), so, if we use the values known on the equation above we get that:

m=2,96x10^-5 moles*35,4 g/mole\\ m=1,048x10^-3 g

Remember that these grammes are found in one liter of solution. So, this means we have 1,048x10^-3 g/L. Previously we said that ppm=mg/L, so all that's left to do it to convert grammes to miligrammes:

1 g = 1000 mg

If we multiply both sides by 1,048x10^-3:

1 * 1,048x10^-3 g = 1000 * 1,048x10^-3 mg

1,048x10^-3 g = 1,048 mg

Knowing that this amount of mass was found in one liter, we get that the amount of substance in the solution is:

1,048 mg/L

Knowing that <em>mg/L=ppm</em>, then the concentration of Cl2 in ppm is:

<u>1,048 ppm</u>

4 0
2 years ago
What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p
Ray Of Light [21]

Answer:

A. Volume is unchanged

Explanation:

P_{i} = initial pressure of the gas = 722 torr = 96258.7 pa

P_{f} = final pressure of the gas = 0.950 atm = 96258.75 pa

T_{i} = initial temperature = 32 °F = 272.15 K

T_{f} = final temperature = 273 K

V_{i} = initial volume

V_{f} = final volume

Using the Equation

\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}

Inserting the values

\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}

V_{f} = (1.00312) V_{i}

Hence the volume is unchanged.

4 0
2 years ago
On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
xenn [34]
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of T=31^{\circ}C:
v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
v= \frac{2L}{t}
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}=  804.1 m
6 0
2 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
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