Two pendulums differ in lengths by 22 cm. Th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the lengths of the two pendulums be $L_1 = x$ and $L_2 = x - 22$. Since the time period $T = 2 \pi \sqrt{\frac{L}{g}}$,the frequency $f = \fra

Vedclass · Accepted Answer

$72$ and $50$

Vedclass · Answer

(A) Let the lengths of the two pendulums be $L_1 = x$ and $L_2 = x - 22$. Since the time period $T = 2 \pi \sqrt{\frac{L}{g}}$,the frequency $f = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{g}{L}}$. In a given time $t$,the number of oscillations $n = f 	imes t = \frac{t}{2 \pi} \sqrt{\frac{g}{L}}$. Since $t$ is the same for both,$n \propto \frac{1}{\sqrt{L}}$. Therefore,$n_1 \sqrt{L_1} = n_2 \sqrt{L_2}$. Given $n_1 = 15$ and $n_2 = 18$,we have $15 \sqrt{x} = 18 \sqrt{x - 22}$. Dividing by $3$,we get $5 \sqrt{x} = 6 \sqrt{x - 22}$. Squaring both sides: $25x = 36(x - 22)$. $25x = 36x - 792$. $11x = 792 \Rightarrow x = 72 \ cm$. The lengths are $72 \ cm$ and $72 - 22 = 50 \ cm$.

Two pendulums differ in lengths by $22 \ cm$. They oscillate at the same place such that one of them makes $15 \ oscillations$ and the other makes $18 \ oscillations$ during the same time. The lengths (in $cm$) of the pendulums are

Similar Questions

In a seconds pendulum,the mass of the bob is $30\, g$. If it is replaced by a $90\, g$ mass,then its time period will be .... $s$.

To show that a simple pendulum executes simple harmonic motion,it is necessary to assume that

The bob of a simple pendulum is a spherical hollow ball filled with water. $A$ plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation,till water is coming out,the time period of oscillation would

The average speed of the bob of a simple pendulum oscillating with a small amplitude $A$ and time period $T$ is

Identify the correct statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module