हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

P the Pendulum of a Certain Clock Has Time Period 2.04 S. How Fast Or Slow Does the Clock Run During 24 Hours? - Physics

Advertisements
Advertisements

प्रश्न

The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours?

योग
Advertisements

उत्तर

Given,
Time period of the clock pendulum = 2.04 s

The number of oscillations made by the pendulum in one day is calculated as \[\frac{\text { Number  of  seconds  in  one  day }}{\text {time  period  of  pendulum  in  seconds }} = \frac{24 \times 3600}{2}\] = 43200

In each oscillation, the clock gets slower by (2.04 − 2.00) s, i.e., 0.04 s.
In one day, it is slowed by = 43200 × (0.04)
                                          = 28.8 min
Thus, the clock runs 28.8 minutes slow during 24 hours.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Simple Harmonics Motion - Exercise [पृष्ठ २५५]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 12 Simple Harmonics Motion
Exercise | Q 34 | पृष्ठ २५५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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?


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?


The force acting on a particle moving along X-axis is F = −k(x − vo t) where k is a positive constant. An observer moving at a constant velocity v0 along the X-axis looks at the particle. What kind of motion does he find for the particle?


The distance moved by a particle in simple harmonic motion in one time period is


The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is


The average energy in one time period in simple harmonic motion is


A pendulum clock that keeps correct time on the earth is taken to the moon. It will run


A pendulum clock keeping correct time is taken to high altitudes,


Select the correct statements.
(a) A simple harmonic motion is necessarily periodic.
(b) A simple harmonic motion is necessarily oscillatory.
(c) An oscillatory motion is necessarily periodic.
(d) A periodic motion is necessarily oscillatory.


A particle moves in a circular path with a continuously increasing speed. Its motion is


Which of the following quantities are always zero in a simple harmonic motion?
(a) \[\vec{F} \times \vec{a} .\]

(b) \[\vec{v} \times \vec{r} .\]

(c) \[\vec{a} \times \vec{r} .\]

(d) \[\vec{F} \times \vec{r} .\]


Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance h directly above the tunnel. The motion of the particle as seen from the earth is
(a) simple harmonic
(b) parabolic
(c) on a straight line
(d) periodic


A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a second pendulum at a place where = π2 m/s2.


The angle made by the string of a simple pendulum with the vertical depends on time as \[\theta = \frac{\pi}{90}  \sin  \left[ \left( \pi  s^{- 1} \right)t \right]\] .Find the length of the pendulum if g = π2 m2.


A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period.


A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.


The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on the surface of planet X such that the acceleration of the planet X is n times greater than the Earth is


State the laws of the simple pendulum?


Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is

T = `2π sqrt("R"/"g")`


Consider two simple harmonic motion along the x and y-axis having the same frequencies but different amplitudes as x = A sin (ωt + φ) (along x-axis) and y = B sin ωt (along y-axis). Then show that

`"x"^2/"A"^2 + "y"^2/"B"^2 - (2"xy")/"AB" cos φ = sin^2 φ`

and also discuss the special cases when

  1. φ = 0
  2. φ = π
  3. φ = `π/2`
  4. φ = `π/2` and A = B
  5. φ = `π/4`

Note: when a particle is subjected to two simple harmonic motions at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×