Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5 \[\times\] 10 = 5 kg \[\left( \because g = 10 {ms}^{- 2} \right)\]
Total force exerted on the block = Weight of the block + spring force
Periodic time of spring is given by,
\[T = 2\pi\sqrt{\left( \frac{m}{k} \right)}\]
\[ \Rightarrow 0 . 314 = 2\pi\sqrt{\left( \frac{0 . 5}{k} \right)}\]
\[ \Rightarrow k = 200 N/m\]
∴ The force exerted by the spring on the block \[\left( F \right)\] is,
F = kx = 200.0 × 0.1 = 20 N
Maximum force = F + weight of the block
= 20 + 5 = 25 N
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.
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?
Consider a particle moving in simple harmonic motion according to the equation x = 2.0 cos (50 πt + tan−1 0.75) where x is in centimetre and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the first time? (b) When does he acceleration have its maximum magnitude for the first time? (c) When does the particle come to rest for the second time ?
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 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.
Consider the situation shown in figure . Show that if the blocks are displaced slightly in opposite direction and released, they will execute simple harmonic motion. Calculate the time period.
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 a particle executing S.H.M oscillates with a frequency v, then the kinetic energy of the particle?
When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is ______.
If a body is executing simple harmonic motion and its current displacements is `sqrt3/2` times the amplitude from its mean position, then the ratio between potential energy and kinetic 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?
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 ______.
