Advertisements
Advertisements
प्रश्न
A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives correct time at the equator. If the clock is taken to the poles it will
पर्याय
run slow
run fast
stop working
give correct time
उत्तर
give correct time
Because the time period of a spring-mass system does not depend on the acceleration due to gravity.
APPEARS IN
संबंधित प्रश्न
Assuming the expression for displacement of a particle starting from extreme position, explain graphically the variation of velocity and acceleration w.r.t. time.
A particle executes S.H.M. with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy.
Show variation of displacement, velocity, and acceleration with phase for a particle performing linear S.H.M. graphically, when it starts from the extreme position.
Can simple harmonic motion take place in a non-inertial frame? If yes, should the ratio of the force applied with the displacement be constant?
A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?
Can the potential energy in a simple harmonic motion be negative? Will it be so if we choose zero potential energy at some point other than the mean position?
A hollow sphere filled with water is used as the bob of a pendulum. Assume that the equation for simple pendulum is valid with the distance between the point of suspension and centre of mass of the bob acting as the effective length of the pendulum. If water slowly leaks out of the bob, how will the time period vary?
A block of known mass is suspended from a fixed support through a light spring. Can you find the time period of vertical oscillation only by measuring the extension of the spring when the block is in equilibrium?
A student says that he had applied a force \[F = - k\sqrt{x}\] on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.
The displacement of a particle in simple harmonic motion in one time period is
The distance moved by a particle in simple harmonic motion in one time period is
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
The average energy in one time period in simple harmonic motion is
A pendulum clock keeping correct time is taken to high altitudes,
Which of the following quantities are always negative in a simple harmonic motion?
(a) \[\vec{F} . \vec{a} .\]
(b) \[\vec{v} . \vec{r} .\]
(c) \[\vec{a} . \vec{r} .\]
(d)\[\vec{F} . \vec{r} .\]
Which of the following will change the time period as they are taken to moon?
(a) A simple pendulum
(b) A physical pendulum
(c) A torsional pendulum
(d) A spring-mass system
A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?
A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according to as y = kt2, where y is the vertical distance covered and k = 1 ms−2, its time period becomes T2. Then, T `"T"_1^2/"T"_2^2` is (g = 10 ms−2)
If the inertial mass and gravitational mass of the simple pendulum of length l are not equal, then the time period of the simple pendulum is
A simple harmonic motion is given by, x = 2.4 sin ( 4πt). If distances are expressed in cm and time in seconds, the amplitude and frequency of S.H.M. are respectively,