Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
It is given that :
Spring constant, k = 100 N/m,
Mass of the block, M = 1 kg
Force, F = 10 N
(a) In the equilibrium position,
F = kx
where x is the compression of the spring, and
k is the spring constant.
∴ `x = F/k = 10/100`
= 0.1 m = 10 cm
(b) The blow imparts a speed of 2 ms-1 to the block, towards left.
Potential energy of spring, U = `1/2 kx^2`
Kinetic energy,
`K = 1/2 mv^2`
(c) Time period (T) is given by,
`T = 2pisqrt(M/k)`
`2pi sqrt(1/100) = pi/5` s
(d) Let A be the amplitude.
Amplitude is the distance between the mean and the extreme position.
At the extreme position, the compression of the spring will be (A + x).
As the total energy in S.H.M. remains constant, we can write:
`1/2k(A + x )^2 = 1/2kx^2 + 1/2Mv^2 + FA`
= 2.5 + 10A
∴ 50(A + 0.1)2 = 2.5 + 10A
⇒ 50A2 + 0.5 + 10A = 2.5 + 10A
⇒ 50A2 = 2
⇒ `A^2 = 2/50 = 4/100`
⇒ `A = 2/10 m = 0.2 m = 20 cm`
(e) Potential Energy at the left extreme will be,
`P.E. = 1/2k(A + x)^2`
= `1/2 xx 100 xx (0.1 + 0.2)^2`
= 50 × (0.09) = 4.5 J
(f) Potential Energy at the right extreme is calculated as:
Distance between the two extremes = 2A
`P.E. = 1/2k(A + x)^2 - F(2A)`
= 4.5 − 10(0.4) = 0.5 J
As the work is done by the external force of 10 N, different values of options (b), (e), and (f) do not violate the law of conservation of energy.
APPEARS IN
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.
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 particle having mass 10 g oscillates according to the equation x = (2.0 cm) sin [(100 s−1)t + π/6]. Find (a) the amplitude, the time period and the spring constant. (c) the position, the velocity and the acceleration at t = 0.
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.
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.
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?
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
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.
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 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?
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.
