Answer:
The appropriate alternative is Option B (0.105 sec.).
Explanation:
The given values are:
Elastic modulus,
Y = 9.10 × 10¹⁰ N/m²
Mass,
m = 10.0 kg
Length of rod,
l = 2.00 m
Diameter,
d = 1.00 mm
Now,
⇒ [tex]Keq=\frac{AY}{l}= \frac{\pi D^2 Y}{4l}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{\pi \times 10^{-6}\times 9.1\times 10^{10}}{4\times 2}[/tex]
⇒ [tex]=3.574\times 10^4[/tex]
The time period will be:
⇒ [tex]T=2\pi \sqrt{\frac{m}{Keq} }[/tex]
On substituting the above values, we get
⇒ [tex]=2\pi \sqrt{\frac{10}{3.574\times 10^4}}[/tex]
⇒ [tex]=0.105 \ seconds[/tex]
The time period resulting oscillations will be 0.1005 seconds.
The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
The given data in the problem is;
[tex]\rm \gamma[/tex] is the elastic modulus=9.10 × 10¹⁰ N/m²
m is the mass= 10.0 kg
l is the length of brass rod= 2.00 m
d is the diameter of 1.00 mm
The value of the equivalent stiffness will be;
[tex]\rm K_{eq}= \frac{AY}{l}\\\\ \rm K_{eq}= = \frac{\pi d^2 Y}{4l} \\\\ \rm K_{eq}=\frac{3.14 \times 10^{-6}\times 9.1 \times 10^{10} }{4\times 2 } \\\\ \rm K_{eq}= 3.574 \times 10^4[/tex]
The time period of the oscillation is given by;
[tex]\rm T = 2 \pi \sqrt{\frac{m}{k_{eq}} } \\\\ \rm T = 2 \times 3.14 \sqrt{\frac{10}{3.574 \times 10^4}[/tex]
[tex]\rm T = 0.105 \ sec[/tex]
Hence the time period resulting oscillations will be 0.1005 seconds.
To learn more about the time period of oscillation refer to the link;
https://brainly.com/question/20070798
