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

P a Body of Mass 2 Kg Suspended Through a Vertical Spring Executes Simple Harmonic Motion of Period 4 S. If the Oscillations Are Stopped and the Body Hangs in Equilibrium

Advertisements
Advertisements

प्रश्न

A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium find the potential energy stored in the spring.

बेरीज
Advertisements

उत्तर

It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,

\[T = 2\pi\sqrt{\left( \frac{m}{k} \right)}\]

where k is the spring constant.

On substituting the respective values, we get:

\[\Rightarrow 4 = 2\pi\sqrt{\frac{2}{k}}\] 

\[ \Rightarrow 2 = \pi\sqrt{\frac{2}{k}}\] 

\[ \Rightarrow 4 =  \pi^2 \left( \frac{2}{k} \right)\] 

\[ \Rightarrow k = \frac{2 \pi^2}{4}\] 

\[             = \frac{\pi^2}{2} =   5  N/m\]

As the restoring force is balanced by the weight, we can write:
        mg kx

\[\Rightarrow x = \frac{mg}{k} = \frac{2 \times 10}{5} = 4\]

Potential Energy \[\left( U \right)\] of the spring is,

\[U = \left( \frac{1}{2} \right)k x^2  = \frac{1}{2} \times 5 \times 16\] 

\[     = 5 \times 8 = 40  J\]

shaalaa.com
Energy in Simple Harmonic Motion
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Simple Harmonics Motion - Exercise [पृष्ठ २५२]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 12 Simple Harmonics Motion
Exercise | Q 12 | पृष्ठ २५२

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 particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?


The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.1 N/m. What mass should be attached to the spring?


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.


A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.


The block of mass m1 shown in figure is fastened to the spring and the block of mass m2 is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m1 + m2)g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?


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.


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 ______.


Motion of an oscillating liquid column in a U-tube is ______.


A body is performing S.H.M. Then its ______.

  1. average total energy per cycle is equal to its maximum kinetic energy.
  2. average kinetic energy per cycle is equal to half of its maximum kinetic energy.
  3. mean velocity over a complete cycle is equal to `2/π` times of its π maximum velocity. 
  4. root mean square velocity is times of its maximum velocity `1/sqrt(2)`.

Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.


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.


An object of mass 0.5 kg is executing a simple Harmonic motion. Its amplitude is 5 cm and the time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = `T/4` s starting from the mean position? Assume that the initial phase of the oscillation is zero.


A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = A sin`(pit)/90`. The ratio of kinetic to the potential energy of this particle at t = 210s will be ______.


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×