Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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.
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 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.
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.
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.
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?
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 ______.
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)`.
Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.
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.
