The length of the second's pendulum in a clock is increased to 4 times its initial length. Calculate the numbe

Question 67 of 82

Q. The length of the second's pendulum in a clock is increased to 4 times its initial length. Calculate the number of oscillations completed by the new pendulum in one minute. OR A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate

(a) Angular frequency,

(b) Frequency of vibrations.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019Subjective· 2mImportance★★★★★

82% · 67/82 Questions

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Simple pendulum period depends on $\sqrt{L}$ ; a spring-mass system's angular frequency depends on $\sqrt{k/m}$ .

Option — Pendulum

For a simple pendulum, $T = 2\pi\sqrt{L/g}$ , so $T \propto \sqrt{L}$ .

If the length becomes $4$ times, $L' = 4L$ :

$T' = 2\pi\sqrt{\frac{4L}{g}} = 2\times2\pi\sqrt{\frac{L}{g}} = 2T$

Since the original is a seconds pendulum, $T = 2\ \text{s}$ , so $T' = 4\ \text{s}$ .

Number of oscillations completed in 1 minute (60 s):

$n = \frac{60}{T'} = \frac{60}{4} = 15\ \text{oscillations}$

— OR (alternative) —

Spring-mass system: $m = 1\ \text{kg}$ , $k = 16\ \text{N/m}$ .

(a) Angular frequency: …

Q.The length of the second's pendulum in a clock is increased to 4 times its initial length. Calculate the number of oscillations completed by the new pendulum in one minute. OR A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate