We are given a series circuit with two light bulbs. In this case, the light bulbs act as resistors in series and the total resistance is:
That is the sum of all the resistances in series in the circuit. To determine the voltage we can use Ohm's law:
Where "R" is the total resistance and "I" is the current in the circuit. Replacing we get:
Answer:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
Explanation:
The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>
We have a block-spring system, whose angular frequency () is defined by the following formula:
(1)
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
And the period (), measured in seconds, is determined by the following expression:
(2)
By applying (1) in (2), we get the following formula:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.