Simple Pendulum

The bob of a simple pendulum is a hollow sphere with a small hole in the bottom. It is filled with sand and set into oscillation. Its periodic time (T) is measured at regular intervals. As the sand starts coming out we find that.

A simple pendulum consists of a hollow sphere filled with mercury. As the mercury begins to leak out slowly through a small hole at the bottom, the time period of the pendulum will

A pendulum clock is running slow, In order to correct it, we should

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 sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately:

(Consider mass of the bob $= 20\text{ g}$)

Using a simple pendulum experiment $g$ is determind by measuring its time period $T$. Which of the following plots represent the correct relation between the pendulum length $L$ and time period $T$?

The time period of a simple pendulum inside a stationary lift is $\sqrt{5}$ s. The time period when the lift moves upwards with an acceleration $\frac{g}{4}$ is ($g$ = acceleration due to gravity)

Two simple pendulums having lengths $\ell_1$ and $\ell_2$ with negligible string mass undergo angular displacement $\theta_1$ and $\theta_2$ from their mean positions respectively. If the angular acceleration of both pendulums are same, then which equation is correct?

A simple pendulum of length $l_1$ has periodic time $2.4$ s. Another simple pendulum of length $l_2 < l_1$, has periodic time $1.8$ s. The periodic time of simple pendulum of length $(l_1 - l_2)$ is nearly

Two simple pendulums of lengths $L_1$ and $L_2$ have periodic time $T_1$ and $T_2$ respectively $(T_1 > T_2)$. The time period of the pendulum of length $(L_1 - L_2)$ is

$[(L_1 - L_2) > 60$ cm$]$

Two oscillating simple pendulums with time periods $T$ and $\frac{4T}{3}$ are in phase at a given time. They will be again in phase after an elapse of time

Two simple pendulums, A of mass $M_1$ and length $L_1$, and B of mass $M_2$ and length $L_2.\ M_1 = M_2$ and $L_1 = 2L_2$. If vibrational energies of both are same then which one of the following is correct?

The path length of oscillation of simple pendulum of $1$ m is $8$ cm. Its maximum velocity is (Take $g = \pi^2$ m/s$^{2}$)

If length of oscillating simple pendulum is made $\left(\frac{1}{3}\right)$ times at a place keeping amplitude same then its total energy will be

A simple pendulum of length $L$ and mass of the bob $m$ is oscillating with amplitude $A$. The maximum tension in the string is ($g$ = acceleration due to gravity)

Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum $L$, notes down the data of time taken to complete 30 oscillations as 60 s and hence calculates the length of the simple pendulum as: (Take $\pi^{2} = 9.8$, and $g = 9.8\text{ m/s}^{2}$)

An oscillating pendulum suspended from the roof of a lift which is at rest has time period $T_1$. When lift moves up with acceleration 'a' its time period is $T_2$. When lift moves down with acceleration 'a' its time period is $T_3$. The relation between $T_1,\ T_2$ and $T_3$ is

The ratio of the frequencies of two simple pendulums is $4 : 3$ at the same place. The ratio of their respective lengths is

The length of the simple pendulum is made 3 times the original length. If ' T ' is its original time period, then the new time period will be

At a place, the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping amplitude same then the total energy will be

If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original keeping amplitude same then increase in its total energy at a place will be

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.

Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.

In the light of the above statements, choose the most appropriate answer from the options given below:

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, that of the Earth. The time period of the pendulum remains same on earth and the planet.

Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below:

The time period of a simple pendulum inside a stationary lift is $\sqrt{3}$ second. When the lift moves upwards with an acceleration $g/3$, the time period will be ( $g =$ acceleration due to gravity)

' $P$ ' and ' $Q$ ' are fixed points in same plane and mass ' $m$ ' is tied by string as shown in figure. If the mass is displaced slightly out of this plane and released, it will oscillate with time period $(PQ = 2d, PR = QR = L)(g =$ gravitational acceleration)

Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $\theta_1$ and $\theta_2$, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

Two simple pendulums have first (A) bob of mass ' $M_1$ ' and length ' $L_1$ ', second (B) of mass ' $M_2$ ' and length ' $L_2$ '. $M_1 = M_2$ and $L_1 = 2L_2$. If their total energies are same then the correct statement is

A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is ' T '. With what acceleration ' $a$ ' should lift be accelerated upward in order to reduce the period to ' $T/2$ '? (take ' g ' as acceleration due to gravity)

A simple pendulum has time period ' $T_1$ '. The point of suspension is now moved upward according to equation $y = kt^{2}$ where $k = 1\text{ m/s}^{2}$. If new time period is ' $T_2$ ' then $T_1^{2}/T_2^{2}$ will be ( $g = 10\text{ m/s}^{2}$)

A simple pendulum of length ' $l$ ' has a brass bob attached at its lower end. It's period is ' T '. A steel bob of the same size, having density ' $x$ ' times that of brass, replaces the brass bob. Its length is then so changed that the period becomes ' 2 T '. What is the new length?

The period of a simple pendulum gets doubled when

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:

A simple pendulum of length ' $L$ ' has mass ' $M$ ' and it oscillates freely with amplitude ' $A$ '. At extreme position, its potential energy is

Choose the correct answer. When a point of suspension of pendulum is moved vertically upward with acceleration ' $a$ ', its period of oscillation

Let ' $l_1$ ' be the length of simple pendulum. Its length changes to ' $l_2$ ' to increase the periodic time by $20\%$. The ratio $\frac{l_2}{l_1} =$

A simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2\ (l_1 > l_2)$ has time period $T_2$. Then the time period of the pendulum of length $(l_1 - l_2)$ will be

The bob of a pendulum of length ' $l$ ' is pulled aside from its equilibrium position through an angle ' $\theta$ ' and then released. The bob will then pass through its equilibrium position with speed ' $v$ ', where ' $v$ ' equal to ( $g =$ acceleration due to gravity)

A simple pendulum of length $1\text{ m}$ has a wooden bob of mass $1\text{ kg}$. It is struck by a bullet of mass $10^{-2}\text{ kg}$ moving with a speed of $2 \times 10^{2}\text{ m s}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $g = 10\text{ m/s}^{2}$)

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10\text{ m}$. If it dissipates $10\%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

[Use, $g : 10\text{ m s}^{-2}$]

A simple pendulum doing small oscillations at a place $R$ height above earth surface has time period of $T_1 = 4\text{ s},\ T_2$ would be its time period if it is brought to a point which is at a height $2R$ from earth surface. Choose the correct relation [$R =$ radius of earth]:

A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum

Choose the correct length (L) versus square of the time period ($T^2$) graph for a simple pendulum executing simple harmonic motion.

T is the time period of simple pendulum on the earth's surface. Its time period becomes $x$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:

In a stationary lift, time period of a simple pendulum is '$T$'. The lift starts accelerating downwards with acceleration $\left(\frac{g}{4}\right)$, then the time period of the pendulum will be

The time period of a simple pendulum inside a stationary lift is '$T$'. When the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$, the time period of the pendulum will be

A simple pendulum has a time period '$T$' in air. Its time period when it is completely immersed in a liquid of density one eighth the density of the material of bob is

A simple pendulum oscillating in air has a period of $\sqrt{3}\text{ s}$. If it is completely immersed in non- viscous liquid, having density $\left(\frac{1}{4}\right)$th of the material of the bob, the new period will be :-

The bob of simple pendulum of length '$L$' is released from a position of small angular displacement $\theta$. Its linear displacement at time '$t$' is ( $g =$ acceleration due to gravity)

A simple pendulum of length '$l$' and a bob of mass '$\text{m}$' is executing S.H.M. of small amplitude '$A$'. The maximum tension in the string will be ($g =$ acceleration due to gravity)

A simple pendulum with length $100\text{ cm}$ and bob of mass $250\text{ g}$ is executing S.H.M. of amplitude $10\text{ cm}$. The maximum tension in the string is found to be $\frac{x}{40}\text{ N}$. The value of $x$ is ___________.

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y = A\sin(\pi t + \phi)$, where time is measured in second. The length of pendulum is

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60$^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms${}^{-1}$. (if g = 10 m/s2)

An oscillating simple pendulum of time period $T_0$ is placed in a lift which is accelerating upwards with $2.5\text{ m/s}^2$. The time period of pendulum in the lift will be (if $g = 10\text{ m/s}^2$)

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with $\frac{g}{6}$ vertically upwards then the time period will be:

(Where g = acceleration due to gravity)

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:

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -

(consider radius of earth $R_E = 6400\text{ km}$ and $g$ on earth $10\text{ m/s}^{2}$)

The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by:

The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $10\text{ s}$. If the metallic bob is immersed in water, then the new time period becomes $5\sqrt{x}$ s. The value of $x$ will be ________.

A pendulum clock is running fast. To correct its time, we should

T0 is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $\frac{1}{16}$ times of its initial value, the modified time period is:

A bob of simple pendulum of mass 'm' perform $\text{SHM}$ with amplitude '$A$' and period 'T'. Kinetic energy of pendulum of displacement $x = \frac{A}{2}$ will be

Given below are two statements:

Statement I: A second's pendulum has a time period of 1 second.

Statement II: It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options given below:

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be:

A child is sitting on a swing which performs S.H.M. It has minimum and maximum heights from ground $0.75\text{ m}$ and $2\text{ m}$ respectively. Its maximum speed will be $\left[g = 10\,\frac{\text{m}}{\text{s}^{2}}\right]$

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-

A simple pendulum of length $L$ has mass $m$ and it oscillates freely with amplitude $A$. At extreme position, its potential energy is ($g =$ acceleration due to gravity)

A bob of a simple pendulum has mass $m$ and is oscillating with an amplitude $a$. If the length of the pendulum is $L$, then the maximum tension in the string is $[\cos 0^\circ = 1,\ g =$ acceleration due to gravity]

A simple pendulum of length $l$ has a bob of mass $m$. It executes SHM of small amplitude A. The maximum tension in the string is ($g =$ acceleration due to gravity)

A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be:

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$. If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$, the time period of the pendulum will be

Two pendulums begin to swing simultaneously. The first pendulum makes nine full oscillations when the other makes seven. The ratio of the lengths of the two pendulums is

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2. Then:

A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 1/16 th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by:

The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

In an experiment for determining the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure.

what is the value of g at the place?

Two simple pendulums of length 1 m and 4 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:

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 bob of a simple pendulum executes simple harmonic motion in water with a period $t,$ while the period of oscillation of the bob is $t_0$ in air. Neglecting frictional force of water and given that the density of the bob is $(4/3) \times 1000\text{ kg/m}^{3}.$ What relationship between $t$ and $t_0$ is true

The length of a simple pendulum executing simple harmonic motion is increased by $21\%$. The percentage increase in the time period of the pendulum of increased length is

A child swinging on a swing in sitting position, stands up, then the time period of the swing will

Two masses $M$A and $M$B are hung from two strings of length $l$A and $l$B respectively. They are executing SHM with frequency relation $f$A = 2$f$B, then relation

The bob of simple pendulum having length $l$, is displaced from mean position to an angular position q with respect to vertical. If it is released, then velocity of bob at equilibrium position

The time period of a simple pendulum is $2\text{ s}$. If its length is increased by 4 times, then its period becomes

Two simple pendulums of length $0.5\text{ m}$ and $2.0\text{ m}$ respectively are given small linear displacement in one direction at the same time. They will again be in the same phase when the pendulum of shorter length has completed oscillations

If the length of a simple pendulum is increased by $2\%$, then the time period

A simple pendulum with a bob of mass $m$ oscillates from $A$ to $C$ and back to $A$ such that $PB$ is $H$. If the acceleration due to gravity is $g$, then the velocity of the bob as it passes through $B$ is

A seconds pendulum is mounted in a rocket. Its period of oscillation will decrease when rocket is

The angular velocity and the amplitude of a simple pendulum is $\omega$ and $a$ respectively. At a displacement $x$ from the mean position if its kinetic energy is $T$ and potential energy is $V$, then the ratio of $T$ to $V$ is

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$, then the time period is given by $T = 2\pi\sqrt{(l/g)}$, where $g$ is equal to