Testbook Logo
Exams SuperCoaching Test Series Skill Academy
More
English
Hindi

Question

Download Solution PDF

If a simple pendulum takes 40 seconds to complete 20 oscillations, then the time period of the simple pendulum is

This question was previously asked in
REET Level 2 Mathematics & Science - 24th July 2022 (S3) (Hindi-English-Sanskrit)
Download PDF Attempt Online
View all REET Papers >
  1. 2 second
  2. 0.2 second
  3. 5 second
  4. 0.5 second

Answer (Detailed Solution Below)

Option 1 : 2 second
Free Tests
View all Free tests >
Free
REET CT 1: CDP (Growth and Development)
29.4 K Users
10 Questions 10 Marks 8 Mins
Start Now

Detailed Solution

Download Solution PDF

Concept:

  • The time taken to complete one oscillation is known as the time period of the pendulum.
  • Mathematically, Time Period = Total time is taken / Number of Oscillations

Calculation:

Given,

Total time is taken simple pendulum = 40 sec

Number of oscillations = 20

We know that,

Time Period = Total time taken / Number of Oscillations

⇒ Time taken = 40 / 20 = 2 sec.

Thus, the time period of the simple pendulum is 2 seconds.

Download Solution PDF
Latest REET Updates

Last updated on May 8, 2025

-> Check REET Qualifying Marks 2025 and also know how to calculate your marks here.

-> The REET Result 2024 for Level 1 & Level 2 has been announced on the official website.

-> The REET Final Answer Key 2025 has also been released.

-> The REET 2024 Exam was held on 27th & 28th February 2025.

-> The candidates who qualify for the REET exam will receive a salary range between Rs. 23,700 to Rs. 44,300. 

-> Also, note during probation, the teachers will receive only the basic salary. Candidates must refer to the REET Previous Year Papers and REET Mock Tests to understand the trend of questions for the exam.

More Pendulums Questions

Q1.The time period of a simple pendulum depends on
Q2.If a spring of mass 20 kg has a spring constant of 5 N/m, then its time period is:
Q3.A slow-running pendulum clock can be speeded up by
Q4.If a clock based on oscillating pendulum is taken from the earth to moon, it will
Q5.A mass falls from a height 'h' and its time of fall 't' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'. Then we can say
Q6.If a simple pendulum takes 40 seconds to complete 20 oscillations, then the time period of the simple pendulum is
Q7.Time period of a simple pendulum in a stationary lift is ‘T'. If the lift accelerates with g/6 vertically upwards then the time period will be : (Where g = acceleration due to gravity)
Q8.Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
Q9.Oscillation time of the seconds pendulum
Q10.If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

More Oscillations Questions

Q1.Match the following entries in column-I and column-II with respect to an oscillating spring-block system: Column-I Column-II (a) Mass of the block is doubled (i) Energy of oscillation becomes 4 times (b) Spring constant is made 4 times (ii) Speed of block becomes 2 times (c) Amplitude of oscillations is doubled (iii) P.E. becomes 4 times (d) Angular frequency is doubled (iv) Time period becomes √2 times (e) length of spring is quadrupled (v) Time period becomes 2 times
Q2.The total energy of a particle executing S.H.M. is proportional to
Q3.In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:
Q4.A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ’b’, the correct equivalence will be
Q5.A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = -αx2. The distance at which the particle stops:
Q6.Two electrons each are fixed at a distance ‘2d’. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: (m = mass of charged particle)
Q7.In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:
Q8.When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:
Q9.In figure (A), mass '2 m' is fixed on mass ‘m', which is attached to two springs of spring constant k. In figure (B), mass 'm' is attached to two springs of spring constant 'k' and '2k'. If mass 'm' in (A) and in (B) are displaced by distance 'x' horizontally and then released, then time period T1 and T2 corresponding to (A) and (B) respectively follow the relation.
Q10.The time period of a simple pendulum depends on
Suggested Test Series
View All >
Pedagogy for All Teaching Exams (Paper 1 & 2) - Let's Crack TET!
345 Total Tests with 3 Free Tests
Start Free Test
Social Science (SST) Content for All Teaching Exams - Let's Crack TET!
191 Total Tests with   1 Free Tests
Start Free Test

More Physics Questions

Q1.What is the region around a magnet where the magnetic force is experienced known as?
Q2.What is the angle that is formed between the incident ray and the normal line called?
Q3.If the force acting on the body is zero, then its momentum is -
Q4.A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound gets reflected from a water surface, the wavelength of the reflected sound will be _______. (Speed of sound in air is 340 ms-1)
Q5.What is the SI unit of current?
Q6.Torque acting on electric dipole moment \(\overrightarrow{P}\) placed in uniform electric field \(\overrightarrow{E}\) is 
Q7.Demodulation is essential feature of a
Q8. If the particles are moving with same velocity, then maximum de-Broglie wavelength is for
Q9.Image formed by an objective of a compound microscope is
Q10.Which of the following has smallest wavelength
Important Exams
Super Coaching
Exams
Test Series
Previous Year Papers
Objective Questions
Testbook USA SAT Exam
Testbook USA ACT Exam
US College Admissions Guide
Testbook Edu Solutions Pvt. Ltd.
1st & 2nd Floor, Zion Building,
Plot No. 273, Sector 10, Kharghar,
Navi Mumbai - 410210
[email protected]
Toll Free:1800 833 0800
Office Hours: 10 AM to 7 PM (all 7 days)
Company
About usCareers We are hiringTeach Online on TestbookPartnersMediaSitemap
Products
Test SeriesTestbook PassOnline CoursesOnline VideosPracticeBlogRefer & EarnBooks
Our Apps
Follow us on
Copyright © 2014-2022 Testbook Edu Solutions Pvt. Ltd.: All rights reserved
User PolicyTermsPrivacy
Get Free Access Now
Hot Links： teen patti wink teen patti 100 bonus teen patti apk all teen patti master