Home
Class 11
PHYSICS
A particle suspended from a vertical spr...

A particle suspended from a vertical spring oscillates 10 times per second. At the highest point of oscillation, the spring becomes unstretched. Take `g=pi^(2)m*s^(-2)`
The speed of the particle when the spring is stretched by 0.2 cm is

A

`15.4cm*s^(-1)`

B

`12.8cm*s^(-1)`

C

`10.8cm*s^(-1)`

D

`11.4cm*s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    CHHAYA PUBLICATION|Exercise INTEGER ANSWER TYPE|5 Videos
  • SIMPLE HARMONIC MOTION

    CHHAYA PUBLICATION|Exercise WBCHSE|14 Videos
  • SIMPLE HARMONIC MOTION

    CHHAYA PUBLICATION|Exercise MULTIPLE CORRECT ANSWER TYPE|6 Videos
  • ROTATION OF RIGID BODIES

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|10 Videos
  • STATICS

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|3 Videos

Similar Questions

Explore conceptually related problems

A long spring is stretched by 2 cm. its potential energy is U . If the spring is stretched by 10 cm , its potential energy would be

A particle starts from the origin with a velocity 5 cm/s and moves in a straight line its accelration at time t seconds being (3t^(2)-5t)cm//s^(2) . Find the velocity of the particle and its distance from the origin at the end of 4 seconds.

When a spring is stretched by a distance x, it exerts a force given by F=( -5x-6x^3 ) N. Find the work done when the spring is stretched from 0.1 m to 0.2 m.

A particle of mass 100 g is suspended from the end of a weightless string of length 100 cm and is rotated in a vertical plane. The speed of the particle is 200 cm*s^(-1) when the string makes an angle of theta =60^@ with the verctcal. Determine (i) the tension in the string when theta =60^@ and

A particle of mass 100 g is suspended from the end of a weightless string of length 100 cm and is rotated in a vertical plane. The speed of the particle is 200 cm*s^(-1) when the string makes an angle of theta =60^@ with the verctcal. Determine (ii) the speed of the particle at the lowest position . Acceleration due to gravity = 980 cm *s^(-2) .

A particle of mass m is attached to three identical massless springs of spring constant 'k' as shown in the figure. The time period of vertical oscillation of the particle is

A particle at the end of a spring executes simple harmonic motion with a period t_(1) , while the corresponding period for another spring is t_(2) . If the period of oscillation with the two springs in series is T, then prove that t_(1)^(2)+t_(2)^(2)=T^(2)