A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N/m, oscillating on a frictionless surface. If the block is displaced 3.5 cm from its equilibrium position and released so that its initial velocity is zero, what is its maximum velocity

The maximum velocity of the block is 2.56 m/s.

Explanation:

Given;

mass of the block, m₁ = 45g = 0.045 kg

spring constant, k = 240 N/m

displacement of the block, x = 3.5 cm = 0.035 m

Apply the principle of conservation of energy;

P.E = K.E

¹/₂kx² = ¹/₂m(v-u)²

where;

v is the maximum velocity of the block

u is the initial velocity of the block, u = 0

¹/₂kx² = ¹/₂mv²

kx² = mv²

Therefore, the maximum velocity of the block is 2.56 m/s.

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.