Print my clock's alarm sound
JavaScript (ES7),  55  54 bytes

f=n=&gt;n?f(n-1)+'beep'+[&quot; &quot;[n&gt;50|n%~~(n**=.4)^52%~n]]:''

Try it online!
How?
Given \$1\le n&lt; 50\$, we want to know the number of consecutive beeps that are expected in this part of the sequence. The exact value is given by:
$$\left\lfloor\sqrt{\frac{2n}{5}}+\frac{1}{2}\right\rfloor$$
which is a slightly modified version of A002024.
But in practice, we only need an exact value on the boundaries of the runs of beeps and we can deal with a few off-by-one errors. That's why we instead compute the following approximation:
$$k=\left\lfloor n^{2/5}\right\rfloor$$
We need to insert a space whenever one of the following conditions is satisfied:

\$k=1\$ and \$n\bmod 1=0\$ (the 2nd part being always true)
\$k=2\$ and \$n\bmod 2=1\$
\$k=3\$ and \$n\bmod 3=0\$
\$k=4\$ and \$n\bmod 4=2\$

All the above conditions can be merged into:
$$(n \bmod k) = (52 \bmod (k+1))$$
\$52\$ being the smallest integer \$x&gt;0\$ such that \$x\bmod 3=1\$, \$x\bmod 4=0\$ and \$x\bmod 5=2\$.
We need an additional test for \$n\ge50\$, where all remaining beeps are concatenated together. Otherwise, unwanted spaces would be inserted, starting at \$n=54\$.
Hence the final JS expression:
n &gt; 50 | n % ~~(n **= 0.4) ^ 52 % ~n

which evaluates to 0 when a space must be inserted.

JavaScript (ES7), 55 bytes
A simpler approach using a lookup bit mask.

f=n=&gt;n?f(--n)+'beep'+(0x222222492555F/2**n&amp;1?' ':''):''

Try it online!