Answer:
a) If the mass is doubled, then the period is increased by [tex]\sqrt{2}[/tex]. Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by [tex]\frac{\sqrt{2}}{2}[/tex]. 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 [tex]\frac{\sqrt{2}}{2}[/tex]. Hence, the period of the system is 1.414 seconds.
Explanation:
The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?
We have a block-spring system, whose angular frequency ([tex]\omega[/tex]) is defined by the following formula:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass, measured in kilograms.
And the period ([tex]T[/tex]), measured in seconds, is determined by the following expression:
[tex]T = \frac{2\pi}{\omega}[/tex] (2)
By applying (1) in (2), we get the following formula:
[tex]T = 2\pi\cdot \sqrt{\frac{m}{k} }[/tex]
a) If the mass is doubled, then the period is increased by [tex]\sqrt{2}[/tex]. Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by [tex]\frac{\sqrt{2}}{2}[/tex]. 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 [tex]\frac{\sqrt{2}}{2}[/tex]. Hence, the period of the system is 1.414 seconds.
