Simple Pendulum Numerical of SHM Class-11 Nootan ISC Physics Ch-23 Simple Harmonic Motion. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Simple Pendulum Numerical of SHM Class-11 Nootan ISC Physics Ch-23 Simple Harmonic Motion
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Book | Nootan |
| Chapter-23 | Simple Harmonic Motion. |
| Topics | Numericals on Simple Pendulum of SHM |
| Academic Session | 2025-2026 |
Numericals on Simple Pendulum of SHM
Q-14: The time-period of a simple pendulum is 4 s and its effective length is 4 m. For what length would it make 30 oscillations in 1 minute?
Ans: T1 = 4 s T2 = 60 / 30 = 2 s
l1 = 4 m l2 = ?
T1 / T2 = √l1 / l2
=> 4 / 2 = √4 / l2
=> 2 = 2 / l2
=> l2 = 1 m
Q-15: The value of the acceleration due to gravity at a planet is 1/9th that of the acceleration due to gravity at the earth. If the time-period of a simple pendulum on earth is 1.0 second, find its time-period on the planet.
Ans: T1 / T2 = √g2 / g1
=> 1 / T2 = √g / g x 9 = 1 / 3
=> T2 = 3.0 s
Q-16: Two pendulums of lengths 100 and 110.25 cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
Ans: T1 = 2 π √100 / g and T2 = √110.25 / g
=> T1 / T2 = √100 / 110.25 = 10 / 10.5 = 20 / 21
=> n2 / n1 = 21 / 20
i.e. n2 : n1 = 21 : 20
Q-17: The simple pendulum whose periodic-time is 2 s, is called ‘seconds pendulum’. What will be the length of seconds pendulum on earth? On moon? The value of g on earth is 9.8 m s^-2 and on moon it is 1/6 th of that on earth.
Ans: 2 = 2 π √l1 / g1 = 2 π √l2 / g2
=> l1 / g1 = l2 / g2 => l1 / l2 = g1 / g2
=> l2 = g2 l1 / g1 = g l1 / 6 x g => l1 / 6
again 20 = 2 π √l1 / 9.8
=> l1 = 99.39 x 10^-2 m = 99.39 cm
and l2 = 99.39 / 6 = 16.56 cm
Q-18: A seconds pendulum is is taken to a height where the value of g is 4.36 m s^-2. What will be its new time- period?
Ans: T1 / T2 = √g2 / g1
=> 2 / T2 = √4.36 / 9.8
=> T2 = 3 s
Q-19: A simple pendulum whose length is 20 cm is suspended from the ceiling of a lift which is rising up with an acceleration of 3.0 m s^-2. Calculate the time-period of the pendulum.
Ans: T1 = 2 π √l / g = 2 π √l / g + a
=> 2 x 3.14 x √0.2 / 12.8 = 0.79 s
Q-20: If radius of the earth is 6.4 x 10^6 m, calculate the periodic-time of a simple pendulum of infinite length.
Ans: Time period of pendulum of infinite length
=> T = 2 π √Re / g
=> 2 π √6.4 x 10^6 / 9.8 = 84.6 min
— : End of Simple Pendulum Numerical of SHM Class-11 Nootan ISC Physics Ch-23 Simple Harmonic Motion :–
Return to : – – Nootan Solutions for ISC Physics Class-11
Thanks
Please share with your friends
