Respuesta :
The string is 36 cm long.
Explanation:
Period of simple is given by the equation
[tex]T=2\pi\sqrt{\frac{l}{g}}[/tex]
It will not depend upon angle and mass of ball.
A student with a stopwatch finds that 10 oscillations take 12 s.
So we can find period for 1 oscillation
[tex]\texttt{Period, T = }\frac{12}{10}=1.2s[/tex]
Substituting in equation of period
[tex]T=2\pi\sqrt{\frac{l}{g}}\\\\1.2=2\pi\sqrt{\frac{l}{9.81}}\\\\l=\frac{1.2^2\times 9.81}{4\pi^2}\\\\l=0.36m[/tex]
The string is 0.36 m long.
The string is 36 cm long.
Answer:
The length of the string is [tex]$L=0.357 \mathrm{~m}$[/tex].
Explanation:
Mass of the Ball [tex]=200 \ g[/tex]
Pulled to an angle [tex]=8.0[/tex]°
Ball makes 10 oscillations =[tex]12 \ s[/tex]
Step 1:
The period T of a simple pendulum of length L is given by
[tex]$T=2 \pi \sqrt{\frac{L}{g}}$[/tex]
where [tex]$g=9.80 \mathrm{~m} / \mathrm{s}^{2}$[/tex] is the acceleration due to gravity. Notice that the period of a simple pendulum depends only on its length and on the magnitude of the gravitational constant. It does not depend on the mass of the object hanging at its end or the amplitude of vibration.
Step 2:
Model the ball-string system as a simple pendulum.
Since the ball makes 10 oscillations in [tex]$12.0 \ s[/tex] , the period of one oscillation is:
[tex]$T=\frac{12.0 \mathrm{~s}}{10}$[/tex]
[tex]$=1.20 \mathrm{~s}$[/tex]
Step 3:
Solving Equation for L we obtain:
[tex]$L=\frac{T^{2} g}{4 \pi^{2}}$[/tex]
Substitute numerical values:
[tex]$L=\frac{(1.20 \mathrm{~s})^{2}\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)}{4 \pi^{2}}$[/tex]
[tex]$L=0.357 \mathrm{~m}$[/tex]
Thus we can say the length of the string is[tex]0.357 \mathrm{~m}$.[/tex]
To learn more about length of the String, refer:
- https://brainly.com/question/23333810
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