Pendulum Time Period Formula
The time period ($T$) of a simple pendulum depends on its length ($L$) and the acceleration due to gravity ($g$). The formula is:
$ T = 2\pi\sqrt{\frac{L}{g}} $
Initial Time Period Calculation
Let the initial time period be $T_0$ and the initial length be $L_0$. According to the formula:
$ T_0 = 2\pi\sqrt{\frac{L_0}{g}} $
New Time Period Calculation
The length of the pendulum is reduced to $\frac{1}{16}$ times its initial value. The new length ($L_1$) is:
$ L_1 = \frac{1}{16} L_0 $
The new time period ($T_1$) is calculated using the new length:
$ T_1 = 2\pi\sqrt{\frac{L_1}{g}} $
Substitute $L_1 = \frac{1}{16} L_0$ into the equation:
$ T_1 = 2\pi\sqrt{\frac{\frac{1}{16} L_0}{g}} $
Simplify the expression:
$ T_1 = 2\pi\sqrt{\frac{1}{16}} \sqrt{\frac{L_0}{g}} $
$ T_1 = 2\pi \left(\frac{1}{4}\right) \sqrt{\frac{L_0}{g}} $
$ T_1 = \frac{1}{4} \left( 2\pi\sqrt{\frac{L_0}{g}} \right) $
Since $T_0 = 2\pi\sqrt{\frac{L_0}{g}}$, we can substitute $T_0$ back into the equation for $T_1$:
$ T_1 = \frac{1}{4} T_0 $
Conclusion
When the length of the pendulum is reduced to $\frac{1}{16}$ times its initial value, the modified time period becomes $\frac{1}{4} T_0$. This corresponds to Option B.
Similar Questions
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Important Questions from Oscillations and Waves
- Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2\text{ s}$. When fork $A$ is loaded with wax, the beat frequency is reduced to $4$ beats in $2\text{ s}$. If the original frequency of tuning fork $B$ is $380\text{ Hz}$ then original frequency of tuning fork $A$ is _________ $\text{Hz}$.
- A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ___________ cm. [Assume that the mass of the pendulum remains same.]
- The velocity of sound in air is doubled when the temperature is raised from $0^\circ\text{C}$ to $\alpha^\circ\text{C}$. The value of $\alpha$ is ________.
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