Advertisements
Advertisements
प्रश्न
Answer the following questions:
A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
Advertisements
उत्तर १
The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.
उत्तर २
The wrist watch uses an electronic system or spring system to give the time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.
संबंधित प्रश्न
The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is ______.
If the metal bob of a simple pendulum is replaced by a wooden bob of the same size, then its time period will.....................
- increase
- remain same
- decrease
- first increase and then decrease.
The phase difference between displacement and acceleration of a particle performing S.H.M. is _______.
(A) `pi/2rad`
(B) π rad
(C) 2π rad
(D)`(3pi)/2rad`
let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)
Answer the following questions:
A time period of a particle in SHM depends on the force constant k and mass m of the particle: `T = 2pi sqrt(m/k)` A simple pendulum executes SHM approximately. Why then is the time
The cylindrical piece of the cork of density of base area A and height h floats in a liquid of density `rho_1`. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
`T = 2pi sqrt((hrho)/(rho_1g)`
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
A clock regulated by seconds pendulum, keeps correct time. During summer, length of pendulum increases to 1.005 m. How much will the clock gain or loose in one day?
(g = 9.8 m/s2 and π = 3.142)
Show that, under certain conditions, simple pendulum performs the linear simple harmonic motion.
If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be ______.
A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s2. The period of oscillation is ______.
Which of the following statements is/are true for a simple harmonic oscillator?
- Force acting is directly proportional to displacement from the mean position and opposite to it.
- Motion is periodic.
- Acceleration of the oscillator is constant.
- The velocity is periodic.
Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.
A body of mass m is situated in a potential field U(x) = U0 (1 – cos αx) when U0 and α are constants. Find the time period of small oscillations.
A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then ______.
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 `x/2` times its original time period. Then the value of x is ______.
