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.
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 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.
The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed?
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?
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.
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.
Solve the previous problem if the pulley has a moment of inertia I about its axis and the string does not slip over it.
A rectangle plate of sides a and b is suspended from a ceiling by two parallel string of length L each in Figure . The separation between the string is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.
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.
Discuss in detail the energy in simple harmonic motion.
When a particle executing S.H.M oscillates with a frequency v, then the kinetic energy of the particle?
A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is ______.
A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy 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)`.
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?
