A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 193 m/s and a frequency of 235 Hz . The amplitude of the standing wave at an antinode is 0.380 cm. A) Calculate the amplitude at a point on the string a distance of 18.0 cm from the left-hand end of the string.
B) How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) Calculate the maximum transverse velocity of the string at this point.
D) Calculate the maximum transverse acceleration of the string at this point.

A. We know that amplitude at x is

Asin (kx)

But k= 2πf/v

k= 2*3.132*235/193= 7.65

So A = 0.35*sin( 7.65x 0.18)= 0.00841m

C

Vmax = Amplitude x angular velocity

= 0.0084 x 2πf

= 0.0084* 2*3.142* 235= 12.4m/s

D. Maximum acceleration = omega² x Amplitude

= (2πf)²* 0.00841= 183.40m/s²

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.

your answer is c! hope this : )