Ask question

Ask question

All categories

3 years ago

7

A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is

the period if the amplitude of the motion is doubled

1 answer:

cricket20 [7]3 years ago

5 0

Answer:

<em>The period of the motion will still be equal to T.</em>

<em></em>

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

$T = 2\pi \sqrt{\frac{M}{k} }$

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>

You might be interested in

A car’s momentum is p when it is traveling with a velocity of v. If the velocity of that car doubles, what is the new momentum o

mamaluj [8]

The answer C. 2p. I took the quiz and I got it correct.

5 0

3 years ago

Read 2 more answers

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago

Which are characteristics of mammals? Check all that apply.

Neporo4naja [7]

Answer:

one of the characteristics of a mammal is their several hollow bones another is their three chambered heart and the last is highly developed nervous system

Explanation:

the reason i picked those three is because not all mammals live their life on land and also mammals font have internal fertillization when they are done they take care of their babies and when they grow up they live their own life

6 0

3 years ago

Assume that you have two objects, one with a mass of 6 kg and the other with a mass of 17 kg, each with a charge of −0.027 c and

kirill115 [55]

Ken jenr ewjk hfjek kwe hwlr herw hriehr jkwehrt

3 0

3 years ago

(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How

OLEGan [10]

Answer:

(a) Current flowing through truck battery is 180 A

(b) Time taken in calculator is 333.33 s

Explanation:

(a) Given:

The charge on the truck battery,q = 720 C

Time, t = 4.00 s

Consider I be the current flowing through truck battery.

The relation between current, charge and time is:

I = q/t

Substitute the suitable values in the above equation.

$I=\frac{720}{4}$

I = 180 A

(b) Given:

The charge on the calculator,q = 7.00 C

The current flowing through calculator, I = 0.3 mA = 0.3 x 10⁻³ A

Consider t be the time.

The relation between current, charge and time is:

t = q/I

Substitute the suitable values in the above equation.

$t=\frac{1}{0.3\times10^{-3} }$

I = 333.33 s

8 0

3 years ago