A simple pendulum takes 45 seconds to complete 30 oscillations what is the time period of pendulum

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A simple pendulum consists of a ball (point-mass) m hanging from a (massless) string of length L and fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum may be obtained $$ \tau = I \alpha \qquad \Rightarrow \qquad -mg \sin\theta\; L = mL^2 \; \frac{d^2\theta}{dt^2} $$ and rearranged as $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0 $$ If the amplitude of angular displacement is small enough, so the small angle approximation ($\sin\theta\approx\theta$) holds true, then the equation of motion reduces to the equation of simple harmonic motion $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0 $$ The simple harmonic solution is $$\theta(t) = \theta_o \cos(\omega t) \ , $$ where $\theta_o$ is the initial angular displacement, and $\omega = \sqrt{g/L}$ the natural frequency of the motion. The period of this sytem (time for one oscillation) is $$ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}} . $$

The period of a pendulum does not depend on the mass of the ball, but only on the length of the string. Two pendula with different masses but the same length will have the same period. Two pendula with different lengths will different periods; the pendulum with the longer string will have the longer period.

How many complete oscillations do the blue and brown pendula complete in the time for one complete oscillation of the longer (black) pendulum?

From this information and the definition of the period for a simple pendulum, what is the ratio of lengths for the three pendula?

With the assumption of small angles, the frequency and period of the pendulum are independent of the initial angular displacement amplitude. A pendulum will have the same period regardless of its initial angle. This simple approximation is illustrated in the animation at left. All three pendulums cycle through one complete oscillation in the same amount of time, regardless of the initial angle.

When the angular displacement amplitude of the pendulum is large enough that the small angle approximation no longer holds, then the equation of motion must remain in its nonlinear form $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0 $$ This differential equation does not have a closed form solution, but instead must be solved numerically using a computer. Mathematica numerically solves this differential equation very easily with the built in function NDSolve[ ].

The small angle approximation is valid for initial angular displacements of about 20° or less. If the initial angle is smaller than this amount, then the simple harmonic approximation is sufficient. But, if the angle is larger, then the differences between the small angle approximation and the exact solution quickly become apparent.

In the animation below left, the initial angle is small. The dark blue pendulum is the small angle approximation, and the light blue pendulum (initially hidden behind) is the exact solution. For a small initial angle, it takes a rather large number of oscillations before the difference between the small angle approximation (dark blue) and the exact solution (light blue) begin to noticeable diverge.

In the animation below right, the initial angle is large. The black pendulum is the small angle approximation, and the lighter gray pendulum (initially hidden behind) is the exact solution. For a large initial angle, the difference between the small angle approximation (black) and the exact solution (light gray) becomes apparent almost immediately.

Answer

Verified

Hint: Time period is the total time taken by the body to complete 1 oscillation about its mean position. While frequency is defined as the number of oscillations made by the body about its mean position in 1 sec.

\[T = \dfrac{1}{f}\]

Complete answer:

Time period is the total time taken by the body to complete 1 oscillation about its mean position. While frequency is defined as the number of oscillations made by the body about its mean position in 1 sec. There is a mutual connection between time period and frequency which is given as:

\[T = \dfrac{1}{f}\]

According to the question the body makes 15 oscillations in 30 seconds so going by the definition of time period, the time taken to complete 1 oscillation is

\[T = \dfrac{{30}}{{15}}\sec = 2\sec \]

And going by the definition of frequency, one oscillation is completed in

\[f = \dfrac{{15}}{{30}}Hz = 0.5Hz\]

This can also be done by the formula connecting the two parameters

\[f = \dfrac{1}{T} = \dfrac{1}{2}Hz \\

\therefore f = 0.5\,Hz \]

Hence, the time period is 2 seconds and the frequency is 0.5 Hertz.

Therefore, the correct answer is option C.

Note: There is a saying ‘Work Smart not hard’. So, looking at the options given in the question, we note that the time period is the same in all options i.e. 2 sec. Using only the relation T=1/f, we would have got our answer. Apply this method in other similar questions to save time and energy especially in competitive exams.

This simple pendulum calculator is a tool that will let you calculate the period and frequency of any pendulum in no time. Read on to learn the period of a pendulum equation and use it to solve all of your pendulum swing problems.

We made a video about pendulums! Watch it here:

First of all, a simple pendulum is defined to be a point mass or bob (taking up no space) that is suspended from a weightless string or rod. Such a pendulum moves in a harmonic motion - the oscillations repeat regularly, and kinetic energy is transformed into potential energy, and vice versa.

If you want to calculate the energy of the pendulum, make sure to use our kinetic energy calculator and potential energy calculator.

Diagram of simple pendulum, an ideal model of a pendulum.
(Chetvorno / Public domain)

Surprisingly, for small amplitudes (small angular displacement from the equilibrium position), the pendulum period doesn't depend either on its mass or on the amplitude. It is usually assumed that "small angular displacement" means all angles between -15º and 15º. The formula for the pendulum period is

T = 2π√(L/g)

where:

T is the period of oscillations - time that it takes for the pendulum to complete one full back-and-forth movement;
L is the length of the pendulum (of the string from which the mass is suspended); and
g is the acceleration of gravity. On Earth, this value is equal to 9.80665 m/s² - this is the default value in the simple pendulum calculator.

You can find the frequency of the pendulum as the reciprocal of period:

f = 1/T = 1/2π√(g/L)

Determine the length of the pendulum. For example, it can be equal to 2 m.
Decide a value for the acceleration of gravity. We will use the Earthly figure of 9.80665 m/s², but feel free to check how the pendulum would behave on other planets.
Calculate the period of oscillations according to the formula above: T = 2π√(L/g) = 2π * √(2/9.80665) = 2.837 s.
Find the frequency as the reciprocal of the period: f = 1/T = 0.352 Hz.
You can also let this simple pendulum calculator perform all calculations for you!

For a pendulum with angular displacement higher than 15º, the period also depends on the moment of inertia of the suspended mass. Then, the period of a pendulum equation has the form of:

T = 2π√(I/mgD)

where:

m is the mass of the pendulum;
I is the moment of inertia of the mass; and
D is the distance from the center of mass to the point of suspension.

To calculate the time period of a simple pendulum, follow the given instructions:

Determine the length L of the pendulum.
Divide L by the acceleration due to gravity, i.e., g = 9.8 m/s².
Take the square root of the value from Step 2 and multiply it by 2π.
Congratulations! You have calculated the time period of a simple pendulum.

To determine the acceleration due to gravity using a simple pendulum, proceed as follows:

Measure the time period T for one oscillation using a stopwatch.
Determine the length L of the pendulum.
Divide the length L by the square of the time period T.
Multiply the value obtained from Step 3 by 4π² and you will get the value for acceleration due to gravity, g.

To calculate the length of a simple pendulum, use the formula L = (T/ 2π)²*g. Where T is the time period of the simple pendulum and g is the acceleration due to gravity.

99.36 cm. Using the formula, L = (T/ 2π)²*g, we can determine that the length of a simple pendulum with a time period of 2 seconds is 99.36 cm.