मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

A Rectangle Plate of Sides a and B is Suspended from a Ceiling by Two Parallel String of Length L Each in Figure . the Separation Between the String is D - Physics

Advertisements
Advertisements

प्रश्न

A rectangle plate of sides a and b is suspended from a ceiling by two parallel string of length L each in Figure . The separation between the string is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.

बेरीज
Advertisements

उत्तर

Let m is the mass of rectangular plate and x is the displacement of the rectangular plate.
During the oscillation, the centre of mass does not change.
Driving force \[\left( F \right)\] is given as,
F = mgsin θ
Comparing the above equation with F = ma, we get: 
\[a   =   \frac{F}{m}   =   g\sin\theta\]

For small values of θ, sinθ can be taken as equal to θ.
Thus, the above equation reduces to:

\[a = g\theta = g\left( \frac{x}{L} \right)                                  \left[ \text { Where  g  and  L  are  constant .} \right]\]

It can be seen from the above equation that, a α x.
Hence, the motion is simple harmonic.
Time period of simple harmonic motion \[\left( T \right)\]is given by,

\[T = 2\pi\sqrt{\frac{\text { displacement }}{\text { Acceleration }}}\] 

\[     = 2\pi\sqrt{\frac{x}{gx/L}} = 2\pi\sqrt{\frac{L}{g}}\]

shaalaa.com
Energy in Simple Harmonic Motion
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Simple Harmonics Motion - Exercise [पृष्ठ २५४]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 12 Simple Harmonics Motion
Exercise | Q 26 | पृष्ठ २५४

संबंधित प्रश्‍न

A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and

(f) at 4 cm away from B going towards A.


A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.


The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position.


Find the elastic potential energy stored in each spring shown in figure, when the block is in equilibrium. Also find the time period of vertical oscillation of the block.


Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.


A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.


Find the elastic potential energy stored in each spring shown in figure when the block is in equilibrium. Also find the time period of vertical oscillation of the block.


Discuss in detail the energy in simple harmonic motion.


Show that for a particle executing simple harmonic motion.

  1. the average value of kinetic energy is equal to the average value of potential energy.
  2. average potential energy = average kinetic energy = `1/2` (total energy)

Hint: average kinetic energy = <kinetic energy> = `1/"T" int_0^"T" ("Kinetic energy") "dt"` and

average potential energy = <potential energy> = `1/"T" int_0^"T" ("Potential energy") "dt"`


When a particle executing S.H.M oscillates with a frequency v, then the kinetic energy of the particle? 


If a body is executing simple harmonic motion and its current displacements is `sqrt3/2` times the amplitude from its mean position, then the ratio between potential energy and kinetic energy is:


A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.


A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.


A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.


Draw a graph to show the variation of P.E., K.E. and total energy of a simple harmonic oscillator with displacement.


A mass of 2 kg is attached to the spring of spring constant 50 Nm–1. The block is pulled to a distance of 5 cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at anytime t.


The total energy of a particle, executing simple harmonic motion is ______.

where x is the displacement from the mean position, hence total energy is independent of x.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×