Advertisements
Advertisements
प्रश्न
A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
What is the amplitude of oscillation?
Advertisements
उत्तर
When the support of the hand is removed, the body oscillates about a mean position.
Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.
Loss in PE of the block = `mgx` ......(i)
Where m = mass of the block
The gain in elastic potential energy of the spring = `1/2 kx^2` ......(ii)
As the two are equal, conserving mechanical energy,
We get, `mgx = 1/2kx^2` or `x = (2mg)/k` ......(iii)
Now, the mean position of oscillation will be, when the block is balanced by the spring.
If x' is the extension in that case, then F = + kx'
But `F = mg`
⇒ `mg = + kx^'`
or `x^' = (mg)/k` ......(iv)
Dividing equation (iii) by equation (iv),
`x/x^' = ((2mg)/k)/((mg)/k)` = 2
⇒ `x = 2x^'`
But given x = 4 cm ......(maximum extension from the unstretched position)
∴ `2x^'` = 4
∴ `x^' = 4/2` = 2 cm
But the displacement of mass from the mean position to the position when the spring attains its natural length is equal to the amplitude of the oscillation.
∴ A = x' = 2 cm
Where A = amplitude of the motion.
APPEARS IN
संबंधित प्रश्न
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 maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s2. Find the position(s) of the particle when the speed is 8 cm/s.
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 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.
In following figure k = 100 N/m M = 1 kg and F = 10 N.
- Find the compression of the spring in the equilibrium position.
- A sharp blow by some external agent imparts a speed of 2 m/s to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant.
- Find the time period of the resulting simple harmonic motion.
- Find the amplitude.
- Write the potential energy of the spring when the block is at the left extreme.
- Write the potential energy of the spring when the block is at the right extreme.
The answer of b, e and f are different. Explain why this does not violate the principle of conservation of energy.
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.
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
The springs shown in the figure are all unstretched in the beginning when a man starts pulling the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block.
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.
Show that for a particle executing simple harmonic motion.
- the average value of kinetic energy is equal to the average value of potential energy.
- 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 the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential 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 ______.
- average total energy per cycle is equal to its maximum kinetic energy.
- average kinetic energy per cycle is equal to half of its maximum kinetic energy.
- mean velocity over a complete cycle is equal to `2/π` times of its π maximum velocity.
- root mean square velocity is times of its maximum velocity `1/sqrt(2)`.
Displacement versus time curve for a particle executing S.H.M. is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.
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.
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 ______.
