Differential Equations in Physics

Description: This quiz covers the applications of differential equations in physics, including topics such as motion, heat transfer, and oscillations.
Number of Questions: 5
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Tags: differential equations physics motion heat transfer oscillations
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A particle moves in a straight line with a velocity given by the function (v(t) = 3t^2 - 2t + 1). What is the particle's acceleration at time (t = 2) seconds?

  1. (4)

  2. (6)

  3. (8)

  4. (10)


Correct Option: B
Explanation:

The acceleration of the particle is the derivative of the velocity function with respect to time. So, (a(t) = v'(t) = 6t - 2). At time (t = 2) seconds, the acceleration is (a(2) = 6(2) - 2 = 10) m/s^2.

A mass of 10 kg is attached to a spring with a spring constant of 100 N/m. The mass is pulled 5 cm to the right of its equilibrium position and released. What is the equation of motion for the mass?

  1. (mx'' + kx = 0)

  2. (mx'' - kx = 0)

  3. (mx'' + kx = 10)

  4. (mx'' - kx = 10)


Correct Option: A
Explanation:

The equation of motion for a mass-spring system is given by (mx'' + kx = 0), where (m) is the mass, (k) is the spring constant, and (x) is the displacement from the equilibrium position. In this case, (m = 10) kg, (k = 100) N/m, and the initial displacement is (x_0 = 0.05) m. So, the equation of motion is (10x'' + 100x = 0).

A metal rod of length (L) is heated at one end so that the temperature at a distance (x) from the heated end is given by the function (T(x) = 100 - 20x). What is the rate of heat flow through the rod at a distance (x = 2) meters from the heated end?

  1. (-40) W

  2. (-20) W

  3. (20) W

  4. (40) W


Correct Option: A
Explanation:

The rate of heat flow through the rod is given by the equation (Q = -kA\frac{dT}{dx}), where (k) is the thermal conductivity of the rod, (A) is the cross-sectional area of the rod, and (\frac{dT}{dx}) is the temperature gradient. In this case, (k = 200) W/(m K), (A = 0.01) m^2, and (\frac{dT}{dx} = -20) K/m. So, the rate of heat flow at (x = 2) meters is (Q = -200(0.01)(-20) = -40) W.

A pendulum of length (L) is released from an angle (\theta_0). What is the equation of motion for the pendulum?

  1. (m\frac{d^2\theta}{dt^2} + mg\sin\theta = 0)

  2. (m\frac{d^2\theta}{dt^2} - mg\sin\theta = 0)

  3. (m\frac{d^2\theta}{dt^2} + mg\cos\theta = 0)

  4. (m\frac{d^2\theta}{dt^2} - mg\cos\theta = 0)


Correct Option: A
Explanation:

The equation of motion for a pendulum is given by (m\frac{d^2\theta}{dt^2} + mg\sin\theta = 0), where (m) is the mass of the pendulum bob, (g) is the acceleration due to gravity, (L) is the length of the pendulum, and (\theta) is the angle the pendulum makes with the vertical. This equation is a second-order nonlinear differential equation.

A circuit consists of a resistor of resistance (R), an inductor of inductance (L), and a capacitor of capacitance (C). The charge on the capacitor is given by the function (q(t) = Q_0\cos\omega t), where (Q_0) is the initial charge on the capacitor and (\omega) is the angular frequency of the circuit. What is the equation of motion for the charge on the capacitor?

  1. (LC\frac{d^2q}{dt^2} + RC\frac{dq}{dt} + \frac{1}{C}q = 0)

  2. (LC\frac{d^2q}{dt^2} - RC\frac{dq}{dt} + \frac{1}{C}q = 0)

  3. (LC\frac{d^2q}{dt^2} + RC\frac{dq}{dt} - \frac{1}{C}q = 0)

  4. (LC\frac{d^2q}{dt^2} - RC\frac{dq}{dt} - \frac{1}{C}q = 0)


Correct Option: A
Explanation:

The equation of motion for the charge on the capacitor in an LRC circuit is given by (LC\frac{d^2q}{dt^2} + RC\frac{dq}{dt} + \frac{1}{C}q = 0). This equation is a second-order linear differential equation.

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