Advertisements
Advertisements
प्रश्न
Repeat the previous exercise if the angle between each pair of springs is 120° initially.
Advertisements
उत्तर
As the particle is pushed against the spring C by the distance x, it experiences a force of magnitude kx.
If the angle between each pair of the springs is 120˚ then the net force applied by the springs A and B is given as,
\[\sqrt{\left( \frac{kx}{2} \right)^2 + \left( \frac{kx}{2} \right)^2 + 2\left( \frac{kx}{2} \right)\left( \frac{kx}{2} \right) \cos 120^\circ} = \frac{kx}{2}\]
Total resultant force \[\left( F \right)\] acting on mass m will be,
\[F = kx + \frac{kx}{2} = \frac{3kx}{2} \]
\[ \therefore a = \frac{F}{m} = \frac{3kx}{2m}\]
\[ \Rightarrow \frac{a}{x} = \frac{3k}{2m} = \omega^2 \]
\[ \Rightarrow \omega = \sqrt{\frac{3k}{2m}}\]
\[ \therefore \text { Time period }, T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2m}{3k}}\]
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 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.
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?
The pendulum of a clock is replaced by a spring-mass system with the spring having spring constant 0.1 N/m. What mass should be attached to the spring?
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.
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.
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 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.
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.
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?
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 ______.
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.
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 ______.
