Physics Assignment: Derivatives, Integration and Electrical Circuits

This physics assignment provides solutions to various problems. Part 1 explores the rate of change of voltage and other functions using derivatives. Part 2 focuses on the displacement and velocity of a spring, and also noise, using derivatives. Part 3 involves solving differential equations and analyzing the decay of current in electrical circuits. The assignment covers topics like rate of change, voltage, work done, derivatives, integration, and electrical circuits, providing a comprehensive understanding of physics concepts. The solutions are presented step-by-step, demonstrating the application of relevant formulas and techniques. This assignment is ideal for students seeking to understand and solve complex physics problems, and is available on Desklib for further study and reference.

Part 1:
Solution 1a): Given that the voltage waveform . We know that the rate of
change of function is .
So, rate of change of is
(By using product rule of derivative)
At seconds, the rate of change is
Hence, rate of change is volts per second decrease.
Solution 1b): Given . We know that
So,
Differentiating with respect to x we get
At , the gradient is
Solution 2: The displacement function is .
The velocity of the end of the spring is
At
… (1)
After seconds, the velocity of the end of the spring is
The acceleration of the end of the spring is
Differentiating equation (1) with respect to t we get
After seconds, acceleration of the end of the spring is
Solution 3: the noise of an engine in decibels is given by , where t is
the time in seconds.
Now, the rate of change of noise is .
So,
At time , the rate of change of noise is
Part 2:

Solution 1: Given, the force , where x is the displacement. The work
done is
Solution 2: Given that . Aim is to find E.
Using integration by part method
Where C is constant of integration.
Solution 3: The alternating voltage v is given as
.
The mean value of the voltage over one cycle is
And RMS of the voltage over one cycle is
Part 3: To find the solution of the differential equation with initial
condition .
Now,
Simplify further we get
Since this implies that . Then the required solution is

a): After ,
b): Suppose that the current take t second to decay 50% of its initial value, then
Solution 2: Given that also given that
So,
Since, this implies that . So
After