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3 years ago

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If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration

of the oscillator change? Please show the equations/work.

1 answer:

Nataly_w [17]3 years ago

6 0

Answer:

When we double the angular velocity the maximum acceleration $(a_{max})$ will changes by a factor of 4.

Explanation:

Given the angular frequency $(\omega)$ of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

$a_{max}=A\omega^2$

Now, according to the problem, the angular frequency $(\omega)$ got doubled.

Let us plug $\omega=2\times \omega$ . Then the maximum acceleration will be $a_{max'}$

$a_{max}=A\omega^2$

$a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega$

$a_{max'}=4a_{max}$

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

You might be interested in

What potential difference is dropped over the 60 ohms resistor

timurjin [86]

Voltage = (current) x (resistance).

The current through the 60 ohms resistor is 0.10 A.

V = (0.10 A) x (60 ohms)

V = 6 volts

7 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

4 0

4 years ago

An object moves 20 m east in 30 s and then returns to its starting point taking an additional 50 s. If west is chosen as the pos

disa [49]

The sign associated with the average velocity of the object is negative.

<h3>Average velocity of the object</h3>

The average velocity of the object is calculated as follows;

Average velocity = (v1 + v2)/2

v1 = -20/30 = -0.667 m/s

v2 = 20/50 = 0.4 m/s

Average velocity = (-0.667 + 0.4)/2

Average velocity = -0.134 m/s

Thus, the sign associated with the average velocity of the object is negative.

Learn more about average velocity here: brainly.com/question/6504879

#SPJ1

7 0

2 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago

Could I get some help with this

nika2105 [10]

please give me brainlest!!

the answer is A.

4 0

3 years ago