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

In a Simple Harmonic Motion - Physics

Advertisements
Advertisements

प्रश्न

In a simple harmonic motion

विकल्प

  •  the potential energy is always equal to the kinetic energy

  • the potential energy is never equal to the kinetic energy

  • the average potential energy in any time interval is equal to the average kinetic energy in that time interval

  • the average potential energy in one time period is equal to the average kinetic energy in this period.

MCQ
Advertisements

उत्तर

the average potential energy in one time period is equal to the average kinetic energy in this period.

The kinetic energy of the motion is given as,

\[\frac{1}{2}k A^2 \cos^2 \omega t\]

The potential energy is calculated as,

\[\frac{1}{2}k A^2 \sin^2 \omega t\]
As the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Simple Harmonics Motion - MCQ [पृष्ठ २५२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 12 Simple Harmonics Motion
MCQ | Q 13 | पृष्ठ २५२

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

A particle in S.H.M. has a period of 2 seconds and amplitude of 10 cm. Calculate the acceleration when it is at 4 cm from its positive extreme position.


The average displacement over a period of S.H.M. is ______.

(A = amplitude of S.H.M.)


Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) a = 0.7x

(b) a = –200x2

(c) a = –10x

(d) a = 100x3


A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate:

a) Angular frequency

b) frequency of vibration.


A particle executing simple harmonic motion comes to rest at the extreme positions. Is the resultant force on the particle zero at these positions according to Newton's first law?


A particle executes simple harmonic motion Let P be a point near the mean position and Q be a point near an extreme. The speed of the particle at P is larger than the speed at Q. Still the particle crosses Pand Q equal number of times in a given time interval. Does it make you unhappy?


A platoon of soldiers marches on a road in steps according to the sound of a marching band. The band is stopped and the soldiers are ordered to break the steps while crossing a bridge. Why?


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


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


A particle moves in the X-Y plane according to the equation \[\overrightarrow{r} = \left( \overrightarrow{i} + 2 \overrightarrow{j} \right)A\cos\omega t .\] 

The motion of the particle is
(a) on a straight line
(b) on an ellipse
(c) periodic
(d) simple harmonic


A pendulum clock giving correct time at a place where g = 9.800 m/s2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.


A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s2?


A small block oscillates back and forth on a smooth concave surface of radius R in Figure. Find the time period of small oscillation.


A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.


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 particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45° with the X-axis. The two motions are given by x = x0 sin ωt and s = s0 sin ωt. Find the amplitude of the resultant motion.


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


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.


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, 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×