Home
Class 11
PHYSICS
Amplitude of an SHO is A. When it is at ...

Amplitude of an SHO is A. When it is at a distance y from the mean position of the path of its oscillation, the SHO receives blow in the direction of its motion, which doubles its velocity instantaneously. Find the new amplitude of its oscillations.

Text Solution

Verified by Experts

Velocity of SHO at displacement y.
`v= omega_(0) sqrt(A^(2) - y^(2))" ""……."(1)`
Now instantaneous velocity after giving blow in the direction of motion is `v_(1)`,
`v= omega_(0) sqrt(A^(2) - y^(2))` where `A_(1)` is new amplitude
`2v= omega_(0) sqrt(A^(2) - y^(2))" "[therefore v_(1)= 2v]`
`therefore 2 omega_(0) sqrt(A^(2) - y^(2)) = omega_(0) sqrt(A_(1)^(2) - y^(2))" "[=therefore " From eq. (1) "]`
`therefore 4 (A^(2)- y^(2))= A_(1)^(2) -y^(2)" "[therefore " Squaring "]`
`therefore 4A^(2) - 4y^(2) = A_(1)^(2) -y^(2)`
`therefore 4A^(2)- 3y^(2) = A_(1)^(2)`
`therefore A_(1) = sqrt(4A^(2)- 3y^(2))`.
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    KUMAR PRAKASHAN|Exercise SECTION-C (OBJECTIVE QUESTIONS)|60 Videos

  • OSCILLATIONS

    KUMAR PRAKASHAN|Exercise SECTION-C (OBJECTIVE QUESTIONS TRUE OR FALSE)|8 Videos

  • OSCILLATIONS

    KUMAR PRAKASHAN|Exercise SECTION-B (ADDITIONAL EXERCISE)|9 Videos

  • OBJECTIVE QUESTIONS AS PER NEW PAPER STYLE

    KUMAR PRAKASHAN|Exercise CHAPTER - 8 (Match Type questions)|5 Videos

  • PHYSICAL WORLD

    KUMAR PRAKASHAN|Exercise SECTION-E (QUESTIONS FROM MODULE)|9 Videos

Similar Questions

Explore conceptually related problems

A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is 1 cm, the magnitude of its acceleration is equal to that of its velocity. Find the time period, maximum velocity and maximum acceleration of SHM.

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

SHO of periodic time 2 second starts its oscillation from the lower end of its path of motion, its phase will be……….at time 2 second.

A particle executes simple harmonic with an amplitude of 5 cm. When the particle is at 4 cm from the mean position the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is

A particle performs a SHM with amplitude of 0.01m and the frequency of its oscillation is 60 Hz. Find the maximum acceleration of a particle.

Find the locus of a point which moves such that its distance from the origin is three times its distance from x-axis.

A mass m is undergoing SHM in the verticl direction about the mean position y_(0) with amplitude A and anglular frequency omega . At a distance y form the mean position, the mass detached from the spring. Assume that th spring contracts and does not obstruct the motion of m . Find the distance y . (measured from the mean position). such that height h attained by the block is maximum (Aomega^(2) gt g) .

A train approaching a hill at a speed of 40 km//hr sounds a whistle of frequency 580 Hz when it is at a distance of 1km from a hill. A wind with a speed of 40km//hr is blowing in the direction of motion of the train Find (i) the frequency of the whistle as heard by an observer on the hill, (ii) the distance from the hill at which the echo from the hill is heard by the driver and its frequency. (Velocity of sound in air = 1, 200 km//hr )

A particle performs SHM of amplitude A along a straight line. When it is at distance sqrt(3)/2 A from mean position, its kinetic energy gets increased by an amount 1/2momega^(2)A^(2) due to an impulsive force. Then its new amplitude becomes.

A point particle if mass 0.1 kg is executing SHM of amplitude 0.1 m . When the particle passes through the mean position, its kinetic energy is 8 xx 10^(-3)J . Write down the equation of motion of this particle when the initial phase of oscillation is 45^(@) .